🧠:Okay, so I need to find all injective functions f from the real numbers to the real numbers such that for every real number x and every positive integer n, the absolute value of the sum from i=1 to n of i times (f(x + i + 1) - f(f(x + i))) is less than 2016. Hmm, let me try to unpack this step by step.First, let's restate the problem to make sure I understand it. We need injective functions f: R → R. Injective means that if f(a) = f(b), then a = b. So f is one-to-one. The condition given is that for every real number x and every positive integer n, the absolute value of that sum is bounded by 2016. The bound has to hold for all x and all n, which is interesting because n can be any positive integer, no matter how large. But the bound is fixed at 2016. So even as n grows, the sum can't exceed 2016 in absolute value. That seems restrictive.Let me write down the expression:For all x ∈ R and n ∈ N,|Σ_{i=1}^n [i (f(x + i + 1) - f(f(x + i)))]| < 2016.So the sum is telescoping? Maybe not immediately obvious. Let me check for small n. Let's take n = 1. Then the sum is just 1*(f(x + 2) - f(f(x + 1))). So |f(x + 2) - f(f(x + 1))| < 2016. Since this has to hold for all x, maybe that term is bounded? But for n=2, the sum is 1*(f(x+2) - f(f(x+1))) + 2*(f(x+3) - f(f(x+2))). So each term in the sum is multiplied by its index i. The total sum's absolute value is less than 2016 regardless of n. So even for large n, the sum can't grow beyond 2016. That suggests that the terms in the sum must be canceling each other out in some way, otherwise, for large n, the sum would exceed 2016.For example, if each term in the sum were a constant, say c, then the sum would be c*(1 + 2 + ... + n) = c*n(n+1)/2. This would go to infinity as n increases unless c=0. So if the terms in the sum form a telescoping series where most terms cancel out, then the total sum could be bounded even for large n.Alternatively, maybe the terms themselves decay rapidly enough so that even when multiplied by i, their sum remains bounded. But given that f is injective and from R to R, it's not obvious how the terms would decay. Also, the problem says "for every real number x", so x is arbitrary. So the relationship must hold for any shift in x.Let me try to see if there's a telescoping nature here. Suppose that each term i*(f(x + i + 1) - f(f(x + i))) can be written as a difference between consecutive terms. For example, suppose that term i is equal to A_{i+1} - A_i for some sequence A_i. Then the sum from i=1 to n would telescope to A_{n+1} - A_1, and then the absolute value would be |A_{n+1} - A_1| < 2016. So if we can express the summands in such a telescoping manner, then the entire sum would be bounded by the difference of two terms, which would need to be bounded by 2016 for all n. That might suggest that A_{n} is bounded, but n can be arbitrary. Wait, but in the sum up to n, the telescoped result would involve A_{n+1} - A_1. So if A_{n} is a linear function of n, then the difference could grow with n, which would contradict the bound. Therefore, perhaps A_{n} has to be a constant function, so that the difference A_{n+1} - A_1 is constant minus constant, hence zero? Wait, but then the telescoped sum would be zero. So maybe if the sum telescopes to zero, then the absolute value is zero, which is certainly less than 2016. That seems like a possible approach.So perhaps we can structure the terms such that each term i*(f(x + i + 1) - f(f(x + i))) is equal to some telescoping difference. Let me see. Let me try to find a function A(i, x) such that A(i+1, x) - A(i, x) = i*(f(x + i + 1) - f(f(x + i))). Then the sum from i=1 to n would be A(n+1, x) - A(1, x). If we can make A(n+1, x) - A(1, x) bounded by 2016 for all n and x, that might help. But this seems abstract. Alternatively, maybe consider the sum S(n, x) = Σ_{i=1}^n [i (f(x + i + 1) - f(f(x + i)))]. Then, we can express S(n, x) in terms of S(n-1, x) plus the nth term. But perhaps there's a recursive relationship here.Alternatively, let's try to look for a pattern by writing out the first few terms for specific n.For n=1: S(1, x) = 1*(f(x + 2) - f(f(x + 1))) < 2016.For n=2: S(2, x) = [f(x+2) - f(f(x+1))] + 2*[f(x+3) - f(f(x+2))].For n=3: S(3, x) = [f(x+2) - f(f(x+1))] + 2*[f(x+3) - f(f(x+2))] + 3*[f(x+4) - f(f(x+3))].Hmm. Let's see if there is a telescoping effect. Let's reorganize the terms:For n=2: S(2, x) = f(x+2) - f(f(x+1)) + 2f(x+3) - 2f(f(x+2)).For n=3: S(3, x) = f(x+2) - f(f(x+1)) + 2f(x+3) - 2f(f(x+2)) + 3f(x+4) - 3f(f(x+3)).So each term after the first one adds a multiple of f(x + i + 1) and subtracts a multiple of f(f(x + i)). Let's see if we can group the terms. For example, in n=2:= [f(x+2) - 2f(f(x+2))] + [2f(x+3) - f(f(x+1))].Wait, not sure. Alternatively, let's look for telescoping in the coefficients. Let's see:Suppose we write each term as i*f(x + i + 1) - i*f(f(x + i)). Then, for each term, the first part is i*f(x + i + 1), and the second part is -i*f(f(x + i)). Let's see if these can be combined with adjacent terms.For example, in the n=2 case:Term 1: 1*f(x + 2) - 1*f(f(x + 1))Term 2: 2*f(x + 3) - 2*f(f(x + 2))So combining, we have:1*f(x + 2) - 1*f(f(x + 1)) + 2*f(x + 3) - 2*f(f(x + 2))Maybe rearrange:[1*f(x + 2) - 2*f(f(x + 2))] + [2*f(x + 3) - 1*f(f(x + 1))]But this doesn't seem to telescope. Alternatively, let's note that in term 1, we have -1*f(f(x + 1)), and in term 2, we have +2*f(x + 3). If there's a telescoping effect, perhaps the negative term from one term cancels with the positive term of another. But the indices are different. Hmm.Alternatively, maybe express the sum S(n, x) as:Sum_{i=1}^n [i f(x + i + 1)] - Sum_{i=1}^n [i f(f(x + i))]Let me denote the first sum as S1 and the second sum as S2.So S(n, x) = S1 - S2.Now, let's try to manipulate S1 and S2. Let's shift the index in S1. Let's set j = i + 1 in S1. Then, when i=1, j=2, and when i=n, j = n + 1. So S1 = Sum_{j=2}^{n+1} (j - 1) f(x + j). Similarly, S2 remains Sum_{i=1}^n i f(f(x + i)).Therefore, S(n, x) = Sum_{j=2}^{n+1} (j - 1) f(x + j) - Sum_{i=1}^n i f(f(x + i)).But j is just a dummy variable, so we can write:= Sum_{i=2}^{n+1} (i - 1) f(x + i) - Sum_{i=1}^n i f(f(x + i))Now, split the first sum into i=2 to n and i = n+1:= [Sum_{i=2}^n (i - 1) f(x + i)] + n f(x + n + 1) - [Sum_{i=1}^n i f(f(x + i))]Then, separate the second sum into i=1 and i=2 to n:= [Sum_{i=2}^n (i - 1) f(x + i)] + n f(x + n + 1) - [1*f(f(x + 1)) + Sum_{i=2}^n i f(f(x + i))]Now, combine the sums from i=2 to n:= Sum_{i=2}^n [(i - 1) f(x + i) - i f(f(x + i))] + n f(x + n + 1) - f(f(x + 1))So S(n, x) = [Sum_{i=2}^n ((i - 1) f(x + i) - i f(f(x + i)))] + n f(x + n + 1) - f(f(x + 1))Hmm, not sure if this helps. Let me see if there's a way to make this expression telescope. Let's look at the term inside the sum: (i - 1) f(x + i) - i f(f(x + i)). Maybe we can factor this or relate it to previous terms.Alternatively, let's try to see if assuming a particular form for f would satisfy the condition. Since the function is injective, linear functions come to mind. Let's suppose f is linear, so f(x) = a x + b, where a ≠ 0 to ensure injectivity.Let's test this. Suppose f(x) = a x + b. Then, f is injective because a ≠ 0. Let's compute f(x + i + 1) = a(x + i + 1) + b = a x + a i + a + b. Similarly, f(f(x + i)) = f(a(x + i) + b) = a(a(x + i) + b) + b = a^2 (x + i) + a b + b. Then, compute the term inside the sum:i [f(x + i + 1) - f(f(x + i))] = i [ (a x + a i + a + b) - (a^2 x + a^2 i + a b + b) ]Simplify:= i [ a x + a i + a + b - a^2 x - a^2 i - a b - b ]= i [ a x (1 - a) + a i (1 - a) + a (1 - b) ]Wait, let's compute term by term:First term inside the brackets: a x + a i + a + bSecond term: -a^2 x - a^2 i - a b - bCombine like terms:For x terms: a x - a^2 x = a x (1 - a)For i terms: a i - a^2 i = a i (1 - a)Constants: a + b - a b - b = a - a bSo overall:= i [ a (1 - a) x + a (1 - a) i + a (1 - b) ]Factor out a (1 - a):= i [ a (1 - a)(x + i) + a (1 - b) ]Hmm. So each term in the sum is linear in x and i, multiplied by i. Then the entire sum S(n, x) would be the sum from i=1 to n of these terms. Let's compute S(n, x):S(n, x) = Σ_{i=1}^n [ i [ a (1 - a)(x + i) + a (1 - b) ] ]= a (1 - a) Σ_{i=1}^n [i(x + i)] + a (1 - b) Σ_{i=1}^n iCompute each sum:First sum: Σ_{i=1}^n i(x + i) = x Σ_{i=1}^n i + Σ_{i=1}^n i^2 = x (n(n + 1)/2) + (n(n + 1)(2n + 1)/6)Second sum: Σ_{i=1}^n i = n(n + 1)/2So putting it all together:S(n, x) = a (1 - a) [ x (n(n + 1)/2) + (n(n + 1)(2n + 1)/6) ] + a (1 - b) [ n(n + 1)/2 ]Now, the problem states that |S(n, x)| < 2016 for all x and all n. But let's analyze the expression above. For this to hold for all x and n, the coefficients of x and the terms involving n must vanish, otherwise, as x or n grow, the sum would exceed 2016.Looking at the first term involving x:a (1 - a) * x * (n(n + 1)/2 )For this term to be bounded for all x and n, the coefficient of x must be zero. Similarly, the coefficients of the polynomial in n must also be zero. Let's set the coefficients to zero.First, coefficient of x:a (1 - a) * (n(n + 1)/2 ) = 0 for all n. Since n(n + 1)/2 is not zero for n ≥ 1, we must have a (1 - a) = 0. Therefore, a = 0 or a = 1. But a ≠ 0 because f is injective (if a = 0, f would be constant, which is not injective). Therefore, a = 1.Now, with a = 1, substitute back into the expression:S(n, x) = 1 * (1 - 1) [ x (n(n + 1)/2 ) + ... ] + 1 * (1 - b) [ n(n + 1)/2 ]So the first term becomes zero, and we have:S(n, x) = (1 - b) * [ n(n + 1)/2 ]Therefore, |(1 - b) * [ n(n + 1)/2 ]| < 2016 for all n.But n(n + 1)/2 grows quadratically with n. The only way this can be bounded for all n is if the coefficient (1 - b) is zero. Therefore, 1 - b = 0 => b = 1.Therefore, f(x) = 1 * x + 1 = x + 1.Wait, let me check. If a = 1 and b = 1, then f(x) = x + 1. Let's verify if this function satisfies the original condition.Compute f(x + i + 1) = (x + i + 1) + 1 = x + i + 2.Compute f(f(x + i)) = f((x + i) + 1) = (x + i + 1) + 1 = x + i + 2.Therefore, f(x + i + 1) - f(f(x + i)) = (x + i + 2) - (x + i + 2) = 0.Therefore, each term in the sum is i * 0 = 0, so the entire sum is 0, and |0| < 2016 holds. Therefore, f(x) = x + 1 is a solution.But the question says "find all injective functions". So maybe this is the only solution? Let's check.Suppose f(x) = x + c for some constant c. Then, let's see:f(x + i + 1) = x + i + 1 + c.f(f(x + i)) = f(x + i + c) = x + i + c + c = x + i + 2c.Thus, f(x + i + 1) - f(f(x + i)) = (x + i + 1 + c) - (x + i + 2c) = 1 - c.Therefore, each term in the sum is i*(1 - c). Then the sum S(n, x) = (1 - c) * Σ_{i=1}^n i = (1 - c) * n(n + 1)/2. The absolute value of this is |(1 - c)| * n(n + 1)/2. For this to be less than 2016 for all n, we must have (1 - c) = 0, so c = 1. Therefore, f(x) = x + 1 is the only linear function that works.But the problem says "find all injective functions". So maybe there are non-linear solutions? Or is f(x) = x + 1 the only solution?We need to check if non-linear functions can satisfy the condition. Let's suppose there is a non-linear injective function f satisfying the given condition.First, recall that injective functions from R to R must be either strictly increasing or strictly decreasing. Otherwise, they would fail the horizontal line test. So f is strictly monotonic.Given that, let's consider whether the terms in the sum can cancel each other out for a non-linear f. The key is that the sum must telescope or somehow cancel such that the total is bounded for all n.Suppose that for each i, the term i*(f(x + i + 1) - f(f(x + i))) is equal to some telescoping difference. For example, perhaps it's equal to A(x, i + 1) - A(x, i). Then the sum would telescope to A(x, n + 1) - A(x, 1), which must be bounded in absolute value by 2016 for all x and n. If we can set up such a telescoping, then the difference A(x, n + 1) - A(x, 1) must be bounded. But since n can be any positive integer, unless A(x, i) itself is bounded as i increases, the difference could become large. However, if the telescoping difference leads to cancellation, such that A(x, n + 1) - A(x, 1) is always bounded, then perhaps this is possible.Alternatively, if for each i, f(x + i + 1) - f(f(x + i)) = 0, then all terms are zero, and the sum is zero, which satisfies the condition. That's the case for f(x) = x + 1. But could there be another function where these differences are non-zero but sum up in a telescoping way?Suppose we have f(x + i + 1) - f(f(x + i)) = c/i, where c is a constant. Then the term i*(f(x + i + 1) - f(f(x + i))) = c, so the sum would be c*n. For this to be bounded by 2016 for all n, we must have c = 0. Hence, f(x + i + 1) - f(f(x + i)) = 0 for all i, which again leads to f(x + i + 1) = f(f(x + i)) for all i and x.But x is arbitrary. Let's set y = x + i. Then, for any real y, f(y + 1) = f(f(y)). So the functional equation is f(y + 1) = f(f(y)) for all y ∈ R.So if we can find injective functions f: R → R such that f(y + 1) = f(f(y)) for all y ∈ R, then such functions would satisfy the original condition because each term in the sum would be zero, leading the entire sum to be zero. Therefore, these functions would satisfy the given inequality.So now the problem reduces to finding all injective functions f: R → R satisfying f(y + 1) = f(f(y)) for all y ∈ R. Let's analyze this functional equation.Given f is injective, so if f(a) = f(b), then a = b.We need f(y + 1) = f(f(y)) for all y ∈ R.Since f is injective, we can apply f^{-1} to both sides:y + 1 = f(y).Wait, because if f is injective, it's invertible. So applying f^{-1} to both sides:f^{-1}(f(y + 1)) = f^{-1}(f(f(y)))Which simplifies to:y + 1 = f(y)Therefore, f(y) = y + 1 for all y ∈ R.Hence, the only injective solution to the functional equation f(y + 1) = f(f(y)) is f(y) = y + 1.Therefore, the only solution is f(x) = x + 1.Wait, let me check that again. If f is injective and satisfies f(y + 1) = f(f(y)), then applying f^{-1} gives y + 1 = f(y). Therefore, f(y) = y + 1. That seems right.But let's verify. Suppose f(y + 1) = f(f(y)), and f is injective. Then, since f is injective, f(a) = f(b) implies a = b. So from f(y + 1) = f(f(y)), we can conclude that y + 1 = f(y). Therefore, f(y) = y + 1. So this is the only solution.Therefore, the only injective function satisfying the given condition is f(x) = x + 1.But let me check for f(x) = x + c. Suppose f(y) = y + c. Then f(y + 1) = y + 1 + c, and f(f(y)) = f(y + c) = y + c + c = y + 2c. Therefore, setting f(y + 1) = f(f(y)) gives y + 1 + c = y + 2c ⇒ 1 + c = 2c ⇒ c = 1. So indeed, only c = 1 works. Therefore, f(x) = x + 1 is the only linear solution.But what if f is non-linear? Suppose f is non-linear but satisfies f(y + 1) = f(f(y)) for all y. For example, let me think of a function that shifts by 1 each time. Wait, but if f(y + 1) = f(f(y)), then the function f must satisfy a kind of recursive relation. Suppose f(y) = y + 1, then f(f(y)) = f(y + 1) = y + 2, but f(y + 1) = y + 2, so indeed f(f(y)) = f(y + 1). So that works. What if f is a different function?Suppose f(y) = y + 1 for y ≤ 0, and f(y) = y + 2 for y > 0. Then check if f(y + 1) = f(f(y)). Let's take y = -0.5. Then f(-0.5) = -0.5 + 1 = 0.5. Then f(f(-0.5)) = f(0.5) = 0.5 + 2 = 2.5. On the other hand, f(y + 1) = f(-0.5 + 1) = f(0.5) = 2.5. So in this case, it works. Wait, but hold on. Let me check for another y. Take y = 0.5. Then f(0.5) = 0.5 + 2 = 2.5. Then f(f(0.5)) = f(2.5) = 2.5 + 2 = 4.5. On the other hand, f(y + 1) = f(0.5 + 1) = f(1.5) = 1.5 + 2 = 3.5. So 4.5 ≠ 3.5. Therefore, this function does not satisfy the equation. Therefore, piecewise functions may not work unless the pieces align perfectly.Alternatively, suppose f is periodic? Wait, but f is injective, so periodic functions can't be injective on R unless the period is zero, which would make it constant, but constants aren't injective. So periodic functions are out.What about exponential functions? Suppose f(y) = e^y. Then f(y + 1) = e^{y + 1}, and f(f(y)) = e^{e^y}. These are not equal, so that doesn't work. Similarly, logarithmic functions won't work because they aren't defined on all R.Another idea: Maybe f is linear plus periodic? But again, injectivity would require the periodic part to be zero.Alternatively, maybe f is affine linear, but we already saw that f(x) = x + 1 is the only solution.Alternatively, suppose f is a translation function, but non-linear. For example, f(y) = y + g(y), where g is some non-linear function. Then f(y + 1) = y + 1 + g(y + 1), and f(f(y)) = f(y + g(y)) = y + g(y) + g(y + g(y)). Therefore, the functional equation is y + 1 + g(y + 1) = y + g(y) + g(y + g(y)). Simplifying, we get 1 + g(y + 1) = g(y) + g(y + g(y)). This seems complicated, but if we suppose that g is a constant function, say g(y) = c, then f(y) = y + c. Then we have 1 + c = c + c ⇒ 1 + c = 2c ⇒ c = 1, which brings us back to the linear solution. Therefore, no non-constant g(y) would satisfy the equation unless it's specifically designed, but given the injectivity requirement, it's hard to see how such a g could exist.Alternatively, suppose that f is an additive function, meaning f(a + b) = f(a) + f(b). But additive functions on R are linear, so f(x) = kx, but we already saw that k must be 1, leading to f(x) = x + c. But additive functions with f(x) = kx require c = 0, but then f(x + 1) = k(x + 1) and f(f(x)) = k(kx) = k^2 x. Setting these equal gives kx + k = k^2 x. For all x, this requires k = k^2 and k = 0. But k = 0 is not injective. Hence, additive functions other than f(x) = x + 1 don't work.Another angle: Suppose f is a strictly increasing function. Then, since it's injective, it must be strictly increasing or decreasing. Let's assume strictly increasing. Then, the functional equation f(y + 1) = f(f(y)) implies that applying f twice is the same as shifting by 1. So f(f(y)) = f(y) + 1? Wait, no. Wait, f(f(y)) = f(y + 1). If f is strictly increasing, then perhaps f(y) = y + 1 is forced. Let me see.Suppose that f is strictly increasing and satisfies f(y + 1) = f(f(y)). Let me consider iterating f. Let's define f^1(y) = f(y), f^2(y) = f(f(y)), and so on. Then the equation becomes f(y + 1) = f^2(y). If we iterate this, perhaps we can find a pattern.For example, f(y + 1) = f^2(y)Then, f(y + 2) = f(f(y + 1)) = f(f^2(y)) = f^3(y)Similarly, f(y + n) = f^{n + 1}(y)But if f is linear, say f(y) = y + 1, then f^{n}(y) = y + n. So f(y + n) = y + n + 1 = f^{n + 1}(y). Which holds. But for a non-linear function, this might not hold.Alternatively, suppose f(y) = y + 1. Then indeed, the functional equation holds. If we suppose another function, say f(y) = y + c, then as before, c must be 1.Alternatively, suppose f is a translation function with varying translation, but injectivity requires it to be strictly increasing. If f(y + 1) = f(f(y)), then the function must "accelerate" in such a way that the shift by 1 in the argument corresponds to a shift in the function's value. But it's unclear how to construct such a function without it being linear.Alternatively, consider that if f is invertible, then we can write f^{-1}(f(y + 1)) = f(y). Therefore, y + 1 = f(y). Wait, this is the same as before. Since f is injective, it's invertible, so applying f^{-1} to both sides of f(y + 1) = f(f(y)) gives y + 1 = f(y). Therefore, f(y) = y + 1 for all y.Therefore, regardless of the approach, the only injective solution is f(y) = y + 1. Hence, this must be the only function that satisfies the given condition.To recap, the key steps were:1. Recognize that the given sum must be bounded for all n and x, which forces the terms to cancel or be zero.2. Testing linear functions and finding that only f(x) = x + 1 works.3. Considering the functional equation f(y + 1) = f(f(y)) derived from setting each term in the sum to zero.4. Using the injectivity of f to deduce that f must satisfy y + 1 = f(y), leading to f(y) = y + 1.5. Confirming that non-linear or other functions cannot satisfy the functional equation without violating injectivity or the boundedness condition.Therefore, the only injective function satisfying the problem's condition is f(x) = x + 1.Final AnswerThe only injective function is boxed{f(x) = x + 1}.

✅:To find all injective functions ( f: mathbb{R} rightarrow mathbb{R} ) such that for every real number ( x ) and every positive integer ( n ),[left| sum_{i=1}^n i left( f(x+i+1) - f(f(x+i)) right) right| < 2016,]we start by analyzing the given condition. The key steps are as follows:1. Telescoping Sum Analysis: The given sum must be bounded by 2016 for all ( x ) and ( n ). This implies that the terms in the sum must either cancel out or be zero for the sum to remain bounded as ( n ) grows.2. Testing Linear Functions: Assuming ( f ) is a linear function ( f(x) = ax + b ), we find that the terms in the sum must vanish. This leads to the conclusion that ( a = 1 ) and ( b = 1 ), resulting in ( f(x) = x + 1 ).3. Functional Equation: For the sum to be zero for all ( n ) and ( x ), each term ( f(x+i+1) - f(f(x+i)) ) must be zero. This leads to the functional equation ( f(y + 1) = f(f(y)) ) for all ( y in mathbb{R} ).4. Injectivity and Functional Equation: Using the injectivity of ( f ), we apply ( f^{-1} ) to both sides of the functional equation ( f(y + 1) = f(f(y)) ), leading to ( y + 1 = f(y) ). This confirms that ( f(y) = y + 1 ).5. Verification: Checking the function ( f(x) = x + 1 ) confirms that it satisfies the given condition, as each term in the sum becomes zero, resulting in the entire sum being zero, which is within the bound of 2016.Thus, the only injective function that satisfies the given condition is:[boxed{f(x) = x + 1}]

🔑:1. We start with the given inequality for the function ( f: mathbb{R} rightarrow mathbb{R} ): [ left|sum_{i=1}^n ileft(f(x+i+1)-f(f(x+i))right)right| < 2016 ] This implies: [ -2016 < sum_{i=1}^n ileft(f(x+i+1) - f(f(x+i))right) < 2016 ]2. Consider the sum for ( n+1 ): [ sum_{i=1}^{n+1} ileft(f(x+i+1) - f(f(x+i))right) ] This can be written as: [ sum_{i=1}^n ileft(f(x+i+1) - f(f(x+i))right) + (n+1)left(f(x+n+2) - f(f(x+n+1))right) ] Using the given inequality, we have: [ -2016 < sum_{i=1}^n ileft(f(x+i+1) - f(f(x+i))right) < 2016 ] Adding ((n+1)left(f(x+n+2) - f(f(x+n+1))right)) to each part of the inequality, we get: [ -2016 + (n+1)left(f(x+n+2) - f(f(x+n+1))right) < sum_{i=1}^{n+1} ileft(f(x+i+1) - f(f(x+i))right) < 2016 + (n+1)left(f(x+n+2) - f(f(x+n+1))right) ] Since the sum (sum_{i=1}^{n+1} ileft(f(x+i+1) - f(f(x+i))right)) must also satisfy the original inequality, we have: [ -2016 < -2016 + (n+1)left(f(x+n+2) - f(f(x+n+1))right) < 2016 ] and [ -2016 < 2016 + (n+1)left(f(x+n+2) - f(f(x+n+1))right) < 2016 ]3. Simplifying these inequalities, we get: [ -4032 < (n+1)left(f(x+n+2) - f(f(x+n+1))right) < 4032 ] for all ( x in mathbb{R} ) and ( n in mathbb{Z}^+ ).4. Suppose there exists ( t ) such that ( f(t+1) - f(f(t)) neq 0 ). For sufficiently large ( m in mathbb{Z}^+ ), we have: [ |(m+1)(f((t-m-1)+m+2) - f(f((t-m-1)+m+1)))| = |(m+1)(f(t+1) - f(f(t)))| > 4032 ] This contradicts the inequality ( -4032 < (m+1)(f(t+1) - f(f(t))) < 4032 ). Therefore, we must have: [ f(t+1) = f(f(t)) ] for all ( t in mathbb{R} ).5. Since ( f ) is injective, ( f(t+1) = f(f(t)) ) implies ( t+1 = f(t) ). Therefore, the function ( f ) must be: [ f(x) = x + 1 ] for all ( x in mathbb{R} ).The final answer is ( boxed{ f(x) = x + 1 } ).

🧠:Okay, let's see. I need to find the radius of a circle that passes through point D and is tangent to AC at point E. The triangle ABC is isosceles with AB = AC = 26 and BC = 20. The altitudes from A and B are AD and BE, respectively. Hmm, let me start by visualizing the triangle.First, since ABC is isosceles with AB = AC, the base is BC = 20. So, vertex A is opposite the base BC. The altitude from A to BC is AD. Similarly, the altitude from B should be BE, which goes to AC. Wait, since ABC is isosceles with AB = AC, the altitude from B would not be on the base, right? Because normally, in an isosceles triangle, the altitude from the apex (which is A here) falls on the base. But the altitude from B would be to the opposite side AC. Let me confirm.Yes, in triangle ABC, the altitude from A is AD, which drops to BC. The altitude from B is BE, which drops to AC. So, points D and E are the feet of these altitudes. Got it.First, I need to find the coordinates of points D and E to work with. Maybe setting up a coordinate system would help here. Let me place point B at (-10, 0) and point C at (10, 0) since BC = 20. Then, since AB = AC = 26, the coordinates of point A can be found. Let's do that.Let me denote the coordinates:- Let’s place BC on the x-axis for simplicity. Let B be at (-10, 0) and C at (10, 0). Then, the midpoint of BC is at (0, 0), but since ABC is isosceles with AB = AC, point A must lie somewhere along the y-axis. Let’s denote A as (0, h). Then, the length AB is 26, so the distance from A(0, h) to B(-10, 0) is sqrt[(-10 - 0)^2 + (0 - h)^2] = sqrt[100 + h^2] = 26. Therefore, 100 + h^2 = 676, so h^2 = 576, which gives h = 24. So, point A is at (0, 24).Alright, so coordinates:- A: (0, 24)- B: (-10, 0)- C: (10, 0)Now, let's find point D, which is the foot of the altitude from A to BC. Since BC is on the x-axis, the altitude from A is vertical, right? Because A is on the y-axis, so the altitude from A to BC is along the y-axis. Therefore, point D is the midpoint of BC, which is (0, 0). Wait, but BC is from (-10, 0) to (10, 0), so midpoint is indeed (0, 0). So, D is (0, 0). But wait, is that correct? Because the altitude from A is perpendicular to BC, which is horizontal, so the altitude is vertical, so it drops from (0,24) straight down to (0,0). So, D is (0,0). Got it.Now, point E is the foot of the altitude from B to AC. Let's find E. First, I need the equation of side AC to find where the altitude from B meets AC.The coordinates of A are (0, 24) and C are (10, 0). The equation of line AC can be found by calculating the slope first. Slope m_AC = (0 - 24)/(10 - 0) = -24/10 = -12/5. So, the equation of AC is y - 24 = -12/5(x - 0), which simplifies to y = -12/5 x + 24.Now, the altitude from B to AC is perpendicular to AC. The slope of AC is -12/5, so the slope of the altitude BE is the negative reciprocal, which is 5/12. Since this altitude passes through point B(-10, 0), its equation is y - 0 = 5/12(x + 10), so y = (5/12)x + 50/12 = (5/12)x + 25/6.Now, the foot of the altitude E is the intersection of this altitude with line AC. So, we need to solve the system:y = -12/5 x + 24 (equation of AC)andy = 5/12 x + 25/6 (equation of altitude BE)Setting them equal:-12/5 x + 24 = 5/12 x + 25/6Multiply both sides by 60 to eliminate denominators:-12/5 x * 60 + 24*60 = 5/12 x *60 + 25/6 *60Simplify:-144x + 1440 = 25x + 250Bring all terms to left side:-144x -25x + 1440 -250 = 0-169x + 1190 = 0So, 169x = 1190x = 1190 / 169Let me compute that. Let's divide 1190 by 169.169*7 = 1183, so 1190 - 1183 = 7. So, x = 7 + 7/169 = 7.04142...Wait, but maybe we can leave it as a fraction: 1190/169. Let me check if 1190 and 169 have a common factor. 169 is 13^2. Let's see 1190 divided by 13: 13*91=1183, 1190-1183=7. So, 1190 = 13*91 +7, so it's not divisible by 13. So, x = 1190/169. Then, y can be found by plugging into one of the equations, say the altitude BE:y = (5/12)x + 25/6So, y = (5/12)(1190/169) + 25/6Let's compute this:First term: (5 * 1190) / (12 * 169) = 5950 / 2028Simplify 5950 and 2028 by dividing numerator and denominator by 2: 2975 / 1014Second term: 25/6 = (25 * 169) / (6 * 169) = 4225 / 1014So, total y = (2975 + 4225)/1014 = 7200/1014Simplify 7200/1014: Divide numerator and denominator by 6: 1200/169So, y = 1200/169Therefore, point E is at (1190/169, 1200/169). Hmm, these fractions are a bit messy, but maybe we can keep them as fractions for exactness.Alternatively, let's check if we can simplify x = 1190/169. Let's see, 1190 divided by 13 is 91.538, which isn't an integer, so maybe it's already in simplest terms. Similarly, y = 1200/169. Okay.So, coordinates of E: (1190/169, 1200/169). Let me note that as (x_E, y_E).Now, the problem states that there is a circle passing through D (which is (0,0)) and tangent to AC at E. We need to find the radius of this circle.When a circle is tangent to a line at a point, the radius at the point of tangency is perpendicular to that line. So, the radius from the center of the circle to E must be perpendicular to AC. Since AC has a slope of -12/5, the radius at E (which is perpendicular) has a slope of 5/12.Therefore, the center of the circle lies along the line perpendicular to AC at E. Let's denote the center as O. So, O lies along the line with slope 5/12 passing through E. Moreover, since the circle passes through D (0,0), the distance from O to D must equal the radius, which is also the distance from O to E (since E is on the circle). Therefore, O lies on the perpendicular to AC at E and is equidistant from E and D.So, let's denote the center O as (h, k). Since O lies on the line perpendicular to AC at E, we can express h and k in terms of a parameter t. Let me find the parametric equations for the line perpendicular to AC at E.The direction vector of the perpendicular is (5, 12) because the slope is 5/12. So, starting from E, moving along this direction, the parametric equations are:h = x_E + 5tk = y_E + 12tSo, center O is (x_E + 5t, y_E + 12t)Since the distance from O to E must equal the distance from O to D.Wait, no. The radius is the distance from O to E (since the circle is tangent at E, so OE is a radius), and also, since D is on the circle, the distance from O to D must equal the radius. Therefore, OD = OE = radius.But OE is the radius, and OD is also the radius. Therefore, O must satisfy OD = OE. So, the center O is equidistant from D and E, and lies on the perpendicular to AC at E. Therefore, the set of points equidistant from D and E is the perpendicular bisector of DE. But O also lies on the perpendicular to AC at E. Therefore, the center O is the intersection of the perpendicular bisector of DE and the perpendicular to AC at E.Alternatively, perhaps we can set up equations based on coordinates.Let me write O as (h, k). Then, since O is on the line perpendicular to AC at E, we have:(k - y_E) = (5/12)(h - x_E)Also, since O is equidistant from E and D:sqrt[(h - x_E)^2 + (k - y_E)^2] = sqrt[(h - 0)^2 + (k - 0)^2]Squaring both sides:(h - x_E)^2 + (k - y_E)^2 = h^2 + k^2Expanding the left side:h^2 - 2h x_E + x_E^2 + k^2 - 2k y_E + y_E^2 = h^2 + k^2Subtract h^2 + k^2 from both sides:-2h x_E + x_E^2 - 2k y_E + y_E^2 = 0Which simplifies to:-2h x_E - 2k y_E + (x_E^2 + y_E^2) = 0Divide both sides by -2:h x_E + k y_E - (x_E^2 + y_E^2)/2 = 0But we also have the equation from the line:k - y_E = (5/12)(h - x_E)So, k = (5/12)(h - x_E) + y_ENow, substitute this into the previous equation:h x_E + [(5/12)(h - x_E) + y_E] y_E - (x_E^2 + y_E^2)/2 = 0This is getting complicated, but let's plug in the values for x_E and y_E.Recall that x_E = 1190/169 and y_E = 1200/169. Let me denote x_E as a and y_E as b for simplicity.So, let a = 1190/169, b = 1200/169.Then, the equation becomes:h a + [(5/12)(h - a) + b] b - (a^2 + b^2)/2 = 0Let me expand this:h a + (5/12)(h - a) b + b^2 - (a^2 + b^2)/2 = 0Combine terms:h a + (5/12) b h - (5/12) a b + b^2 - (a^2)/2 - (b^2)/2 = 0Combine like terms:h [a + (5/12) b] + [ - (5/12) a b + b^2 - (a^2)/2 - (b^2)/2 ] = 0Simplify the constants:First, compute a + (5/12) b:a = 1190/169, b = 1200/169(5/12) b = (5/12)(1200/169) = (5 * 100)/169 = 500/169Thus, a + (5/12) b = (1190 + 500)/169 = 1690/169 = 10Interesting, that's a nice simplification.Now, the other terms:- (5/12) a b + b^2 - (a^2)/2 - (b^2)/2Let me compute each term:First term: - (5/12) a b= - (5/12) * (1190/169) * (1200/169)Second term: + b^2= (1200/169)^2Third term: - (a^2)/2= - (1190/169)^2 / 2Fourth term: - (b^2)/2= - (1200/169)^2 / 2So, combining the second, third, and fourth terms:b^2 - (a^2)/2 - (b^2)/2 = (b^2)/2 - (a^2)/2 = (1/2)(b^2 - a^2)So, the entire constant term is:- (5/12) a b + (1/2)(b^2 - a^2)Let me compute this.First, compute (1/2)(b^2 - a^2):= (1/2)( (1200^2 - 1190^2)/169^2 )Note that 1200^2 - 1190^2 = (1200 - 1190)(1200 + 1190) = 10 * 2390 = 23900Thus, (1/2)(23900)/169^2 = 11950 / 169^2Now, compute - (5/12) a b:a b = (1190 * 1200)/169^2So, - (5/12) a b = - (5/12)(1190 * 1200)/169^2 = - (5 * 1190 * 100)/169^2 = - (595000)/169^2Therefore, the constant term is:-595000 /169^2 + 11950 /169^2 = (-595000 + 11950)/169^2 = (-583,050)/169^2So, combining all terms:h * 10 + (-583050)/169^2 = 0Thus, 10 h = 583050 /169^2Therefore, h = (583050)/(10 * 169^2) = 58305 / (169^2)Compute 58305 divided by 169:169 * 345 = 169*(300 + 45) = 169*300 = 50,700; 169*45 = 7,605; total 50,700 +7,605=58,305. Perfect!So, h = 345 /169Therefore, h = 345/169Then, since k = (5/12)(h - a) + bFirst, compute h - a:h - a = (345/169) - (1190/169) = (345 - 1190)/169 = (-845)/169 = -5Therefore, (5/12)(h - a) = (5/12)(-5) = -25/12Then, k = -25/12 + b = -25/12 + 1200/169Convert to common denominator. Let's compute 1200/169 -25/12:Find LCM of 12 and 169. Since 12 = 2^2*3 and 169=13^2, LCM=12*169=2028So:1200/169 = (1200*12)/2028 = 14,400/202825/12 = (25*169)/2028 = 4,225/2028Thus, k = -4,225/2028 +14,400/2028 = (14,400 -4,225)/2028 =10,175/2028Simplify 10,175 /2028. Let's divide numerator and denominator by GCD(10175,2028). Let's find GCD:Divide 10175 by 2028: 2028*5=10,140, remainder 35.Then GCD(2028,35). 2028 ÷35=57, remainder 35*57=1995, 2028-1995=33. Then GCD(35,33). 35 ÷33=1, remainder 2. GCD(33,2)=1. So GCD is 1. Therefore, 10,175/2028 is in simplest terms.So, center O is at (h, k) = (345/169, 10175/2028)Now, the radius is the distance from O to E (or O to D). Let's compute the distance from O to E:Coordinates of E: (1190/169, 1200/169)Coordinates of O: (345/169, 10175/2028)Wait, let's check the denominators. For k, it's 2028, which is 12*169. Let me convert O's coordinates to have denominator 169 for y-coordinate to make computation easier.Wait, 10175/2028 = (10175 ÷13)/ (2028 ÷13) = 782.692... Hmm, maybe not helpful. Alternatively, let me compute the difference in x and y:Δx = 345/169 -1190/169 = (345 - 1190)/169 = (-845)/169 = -5Δy = 10175/2028 -1200/169 = 10175/2028 - (1200*12)/2028 = 10175/2028 -14400/2028 = (-4225)/2028Therefore, the distance OE is sqrt[ (-5)^2 + (-4225/2028)^2 ] = sqrt[25 + (4225/2028)^2 ]Wait, that seems complicated. Alternatively, perhaps compute using fractions.Wait, but maybe it's easier to compute the distance from O to D, since D is at (0,0):Distance OD: sqrt[(345/169)^2 + (10175/2028)^2 ]Hmm. Alternatively, since we found that h = 345/169, and from the equation 10h = 583050 /169^2, which gave h = 58305 /169^2, which we saw was 345/169. Therefore, perhaps there is a smarter way.Wait, but recall that in the equation earlier, we had:-2h x_E - 2k y_E + (x_E^2 + y_E^2) = 0But once we found h = 345/169 and k = 10175/2028, perhaps plug them in to verify.But maybe instead of computing the distance directly, use the fact that the radius is equal to the distance from O to E, which is equal to the distance from O to D. Alternatively, since we already have coordinates for O, maybe compute OD.Compute OD: sqrt[(345/169)^2 + (10175/2028)^2 ]Let me compute each component:First, (345/169)^2: (345^2)/(169^2)345^2 = (300 +45)^2 = 300^2 +2*300*45 +45^2 =90,000 +27,000 +2,025=119,025So, 119,025 /169^2Second term: (10175/2028)^210175^2: Let's compute that. 10,000^2=100,000,000, 175^2=30,625, and cross term 2*10,000*175=3,500,000. So, (10,000 +175)^2 =100,000,000 +3,500,000 +30,625=103,530,625Denominator: 2028^2= (2028)^2. 2028=12*169, so 2028^2=144*169^2Therefore, (10175/2028)^2=103,530,625/(144*169^2)Therefore, the squared distance OD^2=119,025 /169^2 +103,530,625/(144*169^2)Factor out 1/169^2:[119,025 +103,530,625/144]/169^2Compute numerator:Convert 119,025 to over 144:119,025 =119,025*144/144So:119,025*144 +103,530,625 all over 144Compute 119,025*144:First, 100,000*144=14,400,00019,025*144: 10,000*144=1,440,000; 9,025*144=1,299,600; total 1,440,000 +1,299,600=2,739,600So total 14,400,000 +2,739,600=17,139,600Then add 103,530,625:17,139,600 +103,530,625=120,670,225So numerator is 120,670,225 /144Thus, OD^2= (120,670,225 /144)/169^2 =120,670,225/(144*169^2)Take square root: sqrt(120,670,225)/(sqrt(144)*169)= 10,993.1 / (12*169). Wait, but 10,993.1 squared is 120,670,225?Wait, let me check. 10,000^2=100,000,000. 10,500^2=110,250,000. 11,000^2=121,000,000. So, 10,993^2 is approximately (11,000 -7)^2=121,000,000 -2*11,000*7 +49=121,000,000 -154,000 +49=120,846,049. But that's higher than 120,670,225. So, maybe my calculation is off.Wait, perhaps I made a mistake in computing 10175^2. Let me verify:10175^2:First, compute 10000 + 175 =10175(10000 +175)^2 =10000^2 + 2*10000*175 +175^2=100,000,000 +3,500,000 +30,625=103,530,625. Yes, that's correct.Then 119,025*144: 119,025*100=11,902,500; 119,025*40=4,761,000; 119,025*4=476,100; total 11,902,500 +4,761,000=16,663,500 +476,100=17,139,600. Then 17,139,600 +103,530,625=120,670,225. So that's correct.Then sqrt(120,670,225). Let's see:10,000^2=100,000,00011,000^2=121,000,000So, sqrt(120,670,225) is between 10,990 and 11,000. Let's compute 10,990^2:10,990^2 = (11,000 -10)^2 =121,000,000 -2*11,000*10 +100=121,000,000 -220,000 +100=120,780,100But 120,780,100 is larger than 120,670,225. So, let's try 10,980^2:10,980^2=(10,000+980)^2=10,000^2 +2*10,000*980 +980^2=100,000,000 +19,600,000 +960,400=120,560,400Still less than 120,670,225. Difference: 120,670,225 -120,560,400=109,825Now, 10,980 +x)^2=120,560,400 +2*10,980*x +x^2=120,560,400 +21,960x +x^2. We need this equal to 120,670,225.So, 21,960x +x^2=109,825. Assuming x is small, x≈109,825 /21,960≈5.0. Let's check x=5:21,960*5=109,800. Then 109,800 +25=109,825. Perfect! So x=5.Therefore, sqrt(120,670,225)=10,980 +5=10,985.Therefore, sqrt(120,670,225)=10,985Therefore, OD=10,985/(12*169)Compute 10,985 divided by 12: 10,985 ÷12=915.416...But let's compute 10,985/(12*169)=10,985/(2028)=Divide numerator and denominator by GCD(10,985,2028). Let's see:2028 ÷10,985: 2028*5=10,140; 10,985-10,140=845GCD(2028,845). 2028 ÷845=2 with remainder 2028-1690=338GCD(845,338). 845 ÷338=2 with remainder 845-676=169GCD(338,169)=169. Since 169 divides 338 exactly 2 times. So GCD is 169.Therefore, divide numerator and denominator by 169:10,985 ÷169=652028 ÷169=12Thus, OD=65/12≈5.416666...So, the radius is 65/12. Which is approximately 5.416666...Wait, so the radius is 65/12. Let me check if this makes sense.Alternatively, since we found that OD=65/12, and the problem is asking for the radius, that should be the answer. But let me verify with another approach.Alternatively, since the circle passes through D(0,0) and is tangent to AC at E, then the center lies along the perpendicular to AC at E, which we already used. The radius is the distance from the center to E, which is equal to the distance from the center to D. So, with the center found as (345/169,10175/2028), the radius is 65/12. Let me see if 65/12 reduces or if there's a miscalculation.Wait, 65/12 is approximately 5.416..., which is a reasonable radius. Let me check with coordinates.Wait, the coordinates of O are (345/169,10175/2028). Let me convert these to decimals for better intuition.345 divided by 169: 169*2=338, so 345-338=7. So, 345/169=2 +7/169≈2 +0.0414≈2.041410175 divided by 2028: 2028*5=10,140, 10175-10,140=35. So, 10175/2028≈5 +35/2028≈5 +0.0173≈5.0173So, O is approximately (2.0414,5.0173). Then, distance from O to D(0,0):sqrt(2.0414^2 +5.0173^2)≈sqrt(4.167 +25.173)≈sqrt(29.34)≈5.417, which is 65/12≈5.416666...So, that checks out. Therefore, the radius is 65/12.But let me cross-verify this with another approach to make sure.Alternative approach:We can use the fact that for a circle tangent to line AC at E, the center lies along the line perpendicular to AC at E. Let’s denote the center as O. Then, the radius is OE, which is equal to OD since D is on the circle. So, we need to find a point O on the perpendicular to AC at E such that OD = OE.Alternatively, since O lies on the perpendicular to AC at E, and OD = OE, O is equidistant from D and E and lies on the perpendicular to AC at E. Therefore, O is the intersection point of the perpendicular bisector of DE and the perpendicular to AC at E.Let me find the perpendicular bisector of DE.Coordinates of D: (0,0), E: (1190/169,1200/169)Midpoint of DE: ((1190/169 +0)/2, (1200/169 +0)/2) = (595/169, 600/169)Slope of DE: (1200/169 -0)/(1190/169 -0) = 1200/1190 = 120/119Therefore, the perpendicular bisector of DE has slope -119/120Equation of perpendicular bisector: y - 600/169 = (-119/120)(x -595/169)Now, the center O lies on both this perpendicular bisector and the line perpendicular to AC at E (which has slope 5/12). Let me solve these two equations to find O.Wait, but we already solved this earlier by substitution. But maybe this method is clearer.Equation of line perpendicular to AC at E: y -1200/169 = (5/12)(x -1190/169)Equation of perpendicular bisector of DE: y -600/169 = (-119/120)(x -595/169)We need to solve these two equations.Let me write them as:1) y = (5/12)x - (5/12)(1190/169) + 1200/1692) y = (-119/120)x + (119/120)(595/169) + 600/169Simplify equation 1:Compute (5/12)(1190/169): (5*1190)/(12*169)=5950/2028So, equation 1: y = (5/12)x -5950/2028 +1200/169Convert 1200/169 to denominator 2028: 1200/169 = (1200*12)/2028=14400/2028Thus, equation 1: y = (5/12)x -5950/2028 +14400/2028 = (5/12)x + (14400 -5950)/2028 = (5/12)x +8450/2028Simplify 8450/2028: divide numerator and denominator by 2: 4225/1014. Further divide by 13: 4225 ÷13=325, 1014 ÷13=78. So, 325/78≈4.1667Equation 1: y = (5/12)x +325/78Equation 2: y = (-119/120)x + (119*595)/(120*169) +600/169First, compute (119*595)/(120*169):Calculate numerator: 119*595119*500=59,500119*95=11,305Total:59,500 +11,305=70,805Denominator:120*169=20,280So, term1=70,805/20,280Convert 600/169 to denominator 20,280:600/169= (600*120)/(169*120)=72,000/20,280Therefore, equation 2: y = (-119/120)x +70,805/20,280 +72,000/20,280 = (-119/120)x +142,805/20,280Simplify 142,805/20,280: divide numerator and denominator by 5:28,561/4,056So, equation 2: y = (-119/120)x +28,561/4,056Now, set equation1 = equation2:(5/12)x +325/78 = (-119/120)x +28,561/4,056Multiply all terms by 4,056 (which is LCM of denominators 12,78,120,4056). Wait, 4,056=12*338=12*13*26. Let me check: 12*338=4,056. Yes.Alternatively, LCM of 12,78,120,4056. Let's compute:Prime factors:12=2^2*378=2*3*13120=2^3*3*54056= let's divide by 12: 4056/12=338. Then 338=2*169=2*13^2. So, 4056=2^3*3*13^2Thus LCM is 4056.Multiply all terms by 4056:(5/12)x*4056 +325/78*4056 = (-119/120)x*4056 +28,561/4,056*4056Simplify each term:First term: (5/12)x*4056 =5/12*4056 x=5*338 x=1,690xSecond term:325/78*4056=325*(4056/78)=325*52=16,900Third term:(-119/120)x*4056=(-119/120)*4056 x=(-119)*(4056/120)x=(-119)*(33.8)x= wait, 4056 ÷120=33.8. Hmm, decimal is messy. Alternatively, 4056 divided by 120=4056/(12*10)=338/10=33.8So, -119*33.8= -119*(30 +3.8)= -3,570 -452.2= -4,022.2xFourth term:28,561/4,056*4056=28,561*1=28,561So, equation becomes:1,690x +16,900 = -4,022.2x +28,561Bring all terms to left:1,690x +4,022.2x +16,900 -28,561=0Total x terms:1,690 +4,022.2=5,712.2xConstant term:16,900 -28,561= -11,661Thus:5,712.2x -11,661=0 → x=11,661 /5,712.2≈2.0414Which is approximately 345/169≈2.0414 (since 345 ÷169≈2.0414). So, this matches our previous result.Therefore, x=345/169, then y is calculated as per equation1: y=(5/12)x +325/78Plug x=345/169:y=(5/12)*(345/169) +325/78Compute (5/12)*(345/169)= (1,725)/2,028= simplify divide numerator and denominator by 3: 575/676325/78= (325 ÷13)/(78 ÷13)=25/6≈4.1667Convert 575/676 to over 6 denominator:Wait, 575/676 +25/6. Let's convert to common denominator. LCM of 676 and 6 is 2,028.575/676=575*3/2,028=1,725/2,02825/6=25*338/2,028=8,450/2,028So, y=1,725/2,028 +8,450/2,028=10,175/2,028≈5.0173, which matches previous result.Therefore, the radius is indeed 65/12. So, after multiple methods and verifications, the radius is 65/12.Final Answerboxed{dfrac{65}{12}}

✅:Given triangle (ABC) with (AB = AC = 26) and (BC = 20), we need to find the radius of the circle passing through (D) (the foot of the altitude from (A)) and tangent to (AC) at (E) (the foot of the altitude from (B)).1. Coordinates of Points: - Place (B) at ((-10, 0)), (C) at ((10, 0)), and (A) at ((0, 24)) using the isosceles triangle properties. - (D) is the midpoint of (BC), so (D = (0, 0)).2. Finding Point (E): - The equation of line (AC) is (y = -frac{12}{5}x + 24). - The altitude from (B) to (AC) has a slope of (frac{5}{12}) and passes through (B(-10, 0)), giving the equation (y = frac{5}{12}x + frac{25}{6}). - Solving the intersection of these lines, we find (E = left(frac{1190}{169}, frac{1200}{169}right)).3. Center of the Circle: - The center (O) lies on the line perpendicular to (AC) at (E) with slope (frac{5}{12}). - The center (O) must be equidistant from (D) and (E). Using parametric equations and solving for the coordinates, we find (O = left(frac{345}{169}, frac{10175}{2028}right)).4. Calculating the Radius: - The radius is the distance from (O) to (D). Using the distance formula, we find: [ text{Radius} = sqrt{left(frac{345}{169}right)^2 + left(frac{10175}{2028}right)^2} = frac{65}{12} ]Thus, the radius of the circle passing through (D) and tangent to (AC) at (E) is (boxed{dfrac{65}{12}}).

🔑:1. Identify the given values and properties: - Triangle (ABC) is isosceles with ( |AB| = |AC| = 26 ) and ( |BC| = 20 ). - The altitudes from (A) and (B) intersect the opposite sides at (D) and (E), respectively.2. Determine the height from (A) to (BC): - Since (D) is the midpoint of (BC) (due to symmetry in the isosceles triangle), (BD = DC = 10). - Using the Pythagorean Theorem in (triangle ADB): [ AD = sqrt{AB^2 - BD^2} = sqrt{26^2 - 10^2} = sqrt{676 - 100} = sqrt{576} = 24 ]3. Calculate the area of (triangle ABC): - The area can be calculated using the base (BC) and height (AD): [ text{Area} = frac{1}{2} times BC times AD = frac{1}{2} times 20 times 24 = 240 ]4. Determine the length of (BE): - Since (BE) is an altitude, we use the area to find (BE): [ frac{1}{2} times AC times BE = 240 implies 26 times BE = 480 implies BE = frac{480}{26} = frac{240}{13} ]5. Calculate (EC) using the Pythagorean Theorem in (triangle BEC): - Using (BE = frac{240}{13}) and (BC = 20): [ EC = sqrt{BC^2 - BE^2} = sqrt{20^2 - left(frac{240}{13}right)^2} = sqrt{400 - frac{57600}{169}} = sqrt{frac{67600 - 57600}{169}} = sqrt{frac{10000}{169}} = frac{100}{13} ]6. Apply the Power of a Point theorem from (C): - Let the circle intersect (BC) again at (X): [ left(frac{100}{13}right)^2 = 10 times CX implies frac{10000}{169} = 10 times CX implies CX = frac{1000}{169} ] - Therefore, (XD = 10 - CX = 10 - frac{1000}{169} = frac{1690 - 1000}{169} = frac{690}{169}).7. Apply the Power of a Point theorem from (B): - Let the circle intersect (BE) again at (Y): [ (BD)(BX) = (BY)(BE) implies 10 times left(10 + frac{690}{169}right) = BY times frac{240}{13} ] [ 10 times frac{1690}{169} = BY times frac{240}{13} implies frac{16900}{169} = BY times frac{240}{13} implies 100 = BY times frac{240}{13} implies BY = frac{100 times 13}{240} = frac{1300}{240} = frac{65}{12} ]8. Calculate (YE): - Since (YE = BE - BY): [ YE = frac{240}{13} - frac{65}{12} = frac{240 times 12 - 65 times 13}{13 times 12} = frac{2880 - 845}{156} = frac{2035}{156} = frac{65}{6} ]9. Determine the radius of the circle: - Since the circle is tangent to (AC) at (E), the center lies on the perpendicular to (AC) at (E), which is (YE). Therefore, (YE) is the diameter: [ text{Radius} = frac{YE}{2} = frac{frac{65}{6}}{2} = frac{65}{12} ]The final answer is (boxed{frac{65}{12}})

🧠:Okay, so I need to show that the sum from j=1 to 5 of (1 - t_j) multiplied by the exponential of the sum from k=1 to j of t_k is less than or equal to e^{e^{e^{e}}}. Hmm, that seems pretty complex. Let me try to unpack this step by step.First, let's write out the expression explicitly. The sum is:Σ_{j=1}^5 (1 - t_j) exp(Σ_{k=1}^j t_k)So for each j from 1 to 5, we take (1 - t_j) and multiply it by the exponential of the sum of t_1 up to t_j. Then we sum all those terms. The goal is to prove that this sum is bounded above by e^{e^{e^{e}}}.Since all t_j are real numbers, they can be positive, negative, or zero. The exponential function complicates things because it can grow or decay rapidly depending on the sign of the exponent. So, maybe I need to analyze how each term in the sum contributes and find a maximum possible value.Let me start by expanding the sum term by term. Let's denote S_j = Σ_{k=1}^j t_k. Then the expression becomes:Σ_{j=1}^5 (1 - t_j) e^{S_j}So expanding each term:For j=1: (1 - t_1) e^{t_1}For j=2: (1 - t_2) e^{t_1 + t_2}For j=3: (1 - t_3) e^{t_1 + t_2 + t_3}And so on up to j=5.Now, each term involves (1 - t_j) multiplied by an exponential of the cumulative sum up to t_j. The challenge is to find the maximum of this sum over all real numbers t_1 to t_5.Hmm, this seems like an optimization problem with multiple variables. Maybe I can use calculus and take partial derivatives with respect to each t_j, set them to zero, and solve for the critical points? But with five variables, that might get quite involved. Also, considering that exponentials can lead to nonlinear equations, which might not have closed-form solutions. Alternatively, maybe there's a clever substitution or inequality I can use here.Wait, the problem mentions the upper bound is e^{e^{e^{e}}}, which is a tower of exponentials. That suggests that maybe the maximum of this sum is achieved when each t_j is chosen such that the exponents grow exponentially, leading to the tower. Maybe there's a recursive structure here?Alternatively, perhaps induction could be used? But since the problem is fixed for j=1 to 5, induction might not directly apply. Let me think.Alternatively, maybe we can analyze each term and bound them individually, then multiply the bounds? But since the terms are not independent (they share variables t_j), this might not work. For example, t_1 appears in all terms since S_j includes t_1 for each j ≥ 1. So the variables are nested in each term.Alternatively, maybe consider substituting variables for the cumulative sums. Let me define:S_1 = t_1S_2 = S_1 + t_2S_3 = S_2 + t_3S_4 = S_3 + t_4S_5 = S_4 + t_5So then, t_j = S_j - S_{j-1} for j ≥ 1, where S_0 = 0.Rewriting the original sum in terms of S_j:Σ_{j=1}^5 (1 - (S_j - S_{j-1})) e^{S_j}So each term becomes (1 - S_j + S_{j-1}) e^{S_j}Hmm, maybe this substitution can help. Let's write out each term:For j=1: (1 - S_1 + S_0) e^{S_1} = (1 - S_1 + 0) e^{S_1} = (1 - S_1) e^{S_1}For j=2: (1 - S_2 + S_1) e^{S_2}For j=3: (1 - S_3 + S_2) e^{S_3}For j=4: (1 - S_4 + S_3) e^{S_4}For j=5: (1 - S_5 + S_4) e^{S_5}So the entire sum becomes:(1 - S_1)e^{S_1} + (1 - S_2 + S_1)e^{S_2} + (1 - S_3 + S_2)e^{S_3} + (1 - S_4 + S_3)e^{S_4} + (1 - S_5 + S_4)e^{S_5}Hmm, this seems a bit more structured. Maybe we can telescope this sum?Looking at each term, for example, the term for j=2 is (1 - S_2 + S_1)e^{S_2}. If I consider the terms with S_{j-1} in them, perhaps there's a telescoping effect when summing up.Let me write out all the terms:Term1: (1 - S_1)e^{S_1}Term2: (1 - S_2 + S_1)e^{S_2} = (1 - S_2)e^{S_2} + S_1 e^{S_2}Term3: (1 - S_3 + S_2)e^{S_3} = (1 - S_3)e^{S_3} + S_2 e^{S_3}Term4: (1 - S_4 + S_3)e^{S_4} = (1 - S_4)e^{S_4} + S_3 e^{S_4}Term5: (1 - S_5 + S_4)e^{S_5} = (1 - S_5)e^{S_5} + S_4 e^{S_5}So, the entire sum is:[ (1 - S_1)e^{S_1} + (1 - S_2)e^{S_2} + (1 - S_3)e^{S_3} + (1 - S_4)e^{S_4} + (1 - S_5)e^{S_5} ] +[ S_1 e^{S_2} + S_2 e^{S_3} + S_3 e^{S_4} + S_4 e^{S_5} ]So, we have two separate sums. Let's denote the first part as Sum1 and the second part as Sum2:Sum1 = Σ_{j=1}^5 (1 - S_j) e^{S_j}Sum2 = Σ_{j=1}^4 S_j e^{S_{j+1}}Wait, because in Sum2, the terms are S_1 e^{S_2}, S_2 e^{S_3}, S_3 e^{S_4}, S_4 e^{S_5}, so j from 1 to 4.So, total sum = Sum1 + Sum2.But maybe we can find a recursive relation or a telescoping behavior here.Alternatively, let's consider each pair of terms. For example, looking at Term1 and the first part of Term2:Term1: (1 - S_1)e^{S_1}Term2's Sum1 part: (1 - S_2)e^{S_2}But also, Term2 has S_1 e^{S_2} in Sum2.Hmm, maybe integrating factors or derivatives? For example, the function (1 - S)e^{S} is similar to the derivative of e^{S}, but let's check:d/dS [e^S] = e^SBut (1 - S)e^S is different. Wait, what's the derivative of (1 - S)e^S?Let's compute: d/dS [(1 - S)e^S] = -e^S + (1 - S)e^S = (1 - S - 1)e^S = -S e^S.Hmm, interesting. So the derivative is -S e^S. Not sure if that helps.Alternatively, maybe considering the terms in Sum1 and Sum2 together. Let's look at the combination of Sum1 and Sum2.For example, for each j from 1 to 4, we have in Sum1: (1 - S_j)e^{S_j} and in Sum2: S_j e^{S_{j+1}}.So for each j from 1 to 4, the combined contribution is (1 - S_j)e^{S_j} + S_j e^{S_{j+1}}.Similarly, for j=5, we only have the Sum1 term: (1 - S_5)e^{S_5}.So, maybe grouping these terms:For j=1: (1 - S_1)e^{S_1} + S_1 e^{S_2}For j=2: (1 - S_2)e^{S_2} + S_2 e^{S_3}For j=3: (1 - S_3)e^{S_3} + S_3 e^{S_4}For j=4: (1 - S_4)e^{S_4} + S_4 e^{S_5}For j=5: (1 - S_5)e^{S_5}So, maybe we can analyze each of these grouped terms. Let's take the first group:(1 - S_1)e^{S_1} + S_1 e^{S_2}Similarly, the next group is (1 - S_2)e^{S_2} + S_2 e^{S_3}, and so on.If we can bound each of these groups individually, then the total sum would be bounded by the product or sum of these individual bounds. But how?Alternatively, perhaps consider each group as a function of S_j and S_{j+1}, and try to maximize each term.Wait, for the first group:(1 - S_1)e^{S_1} + S_1 e^{S_2}Assuming that S_2 is related to S_1 via S_2 = S_1 + t_2, but since t_2 can be any real number, S_2 can be any real number greater than S_1 (if t_2 is positive) or less than S_1 (if t_2 is negative). Wait, actually, since t_2 is real, S_2 can be any real number. So for fixed S_1, S_2 can be anything. Therefore, for each group, we might be able to maximize over S_{j+1} given S_j.Wait, for the first group: given S_1, the term is (1 - S_1)e^{S_1} + S_1 e^{S_2}. To maximize this term, we need to choose S_2 (given S_1) such that S_1 e^{S_2} is maximized. But S_2 can be any real number. However, if S_1 is positive, then to maximize S_1 e^{S_2}, we need S_2 to be as large as possible, but since there's no upper bound on S_2, this term can be made arbitrarily large. But that contradicts the problem statement which says the total sum is bounded by e^{e^{e^{e}}}. Therefore, my approach must be flawed.Wait, but S_2 is not entirely independent. Because S_2 = S_1 + t_2, and t_2 is a real number. But if t_2 can be any real number, then S_2 can indeed be any real number, regardless of S_1. So for each term, the subsequent S_{j+1} can be chosen independently? Hmm, but actually, S_{j+1} = S_j + t_{j+1}, so once S_j is fixed, t_{j+1} can be chosen to set S_{j+1} to any real number. Therefore, in effect, each S_{j+1} can be considered as a variable independent of S_j. Wait, but that would mean that the variables S_1, S_2, S_3, S_4, S_5 are all independent variables. Is that the case?Wait, originally, we have t_1, t_2, t_3, t_4, t_5 as real numbers. Then S_1 = t_1, S_2 = t_1 + t_2, S_3 = t_1 + t_2 + t_3, etc. Therefore, S_j = S_{j-1} + t_j. So if we consider S_1, S_2, ..., S_5 as variables, they can be any real numbers with no restrictions, because each t_j can be adjusted to make S_j whatever we want, regardless of previous S_{j-1}. For example, to get S_2, we set t_2 = S_2 - S_1. So yes, S_1, S_2, ..., S_5 can be any real numbers; there are no constraints between them. Therefore, in the expression Sum1 + Sum2, all S_j are independent variables. Wait, that's a crucial point. So even though S_j = S_{j-1} + t_j, since t_j can be any real number, S_j can be set to any real number regardless of S_{j-1}. Therefore, effectively, S_1, S_2, S_3, S_4, S_5 are all independent variables. Therefore, we can treat each S_j as independent variables when maximizing the expression.Therefore, the problem reduces to maximizing the expression:Sum = (1 - S_1)e^{S_1} + (1 - S_2 + S_1)e^{S_2} + (1 - S_3 + S_2)e^{S_3} + (1 - S_4 + S_3)e^{S_4} + (1 - S_5 + S_4)e^{S_5}But given that each S_j is independent. Wait, but in the rewritten expression in terms of S_j, the terms are:Term1: (1 - S_1)e^{S_1}Term2: (1 - S_2 + S_1)e^{S_2}Term3: (1 - S_3 + S_2)e^{S_3}Term4: (1 - S_4 + S_3)e^{S_4}Term5: (1 - S_5 + S_4)e^{S_5}But if S_1, S_2, ..., S_5 are independent variables, then we can choose each S_j to maximize the sum. But then, since variables are independent, we can maximize each term individually. However, some terms share variables. For example, Term2 involves both S_1 and S_2, Term3 involves S_2 and S_3, etc. So the variables are not all independent in the expression. Wait, but if S_1 is chosen first, then Term1 is (1 - S_1)e^{S_1}, and Term2 is (1 - S_2 + S_1)e^{S_2}, which involves S_1 and S_2. So if we fix S_1, then Term2 can be maximized over S_2, but S_1 is already fixed. Similarly, for Term3, which involves S_2 and S_3, if S_2 is fixed from Term2, then Term3 can be maximized over S_3, and so on.Therefore, maybe the maximum can be found by sequentially optimizing each variable, starting from S_1, then S_2, etc., in a backward manner?Alternatively, since the expression is a sum where each term after the first includes variables from the previous terms, perhaps dynamic programming could be useful here. Let me consider optimizing the variables one by one.Let me start with Term5: (1 - S_5 + S_4)e^{S_5}. For fixed S_4, how do we choose S_5 to maximize this term? Let's treat S_4 as a constant and maximize with respect to S_5.Let f(S_5) = (1 - S_5 + S_4)e^{S_5} = (1 + S_4 - S_5)e^{S_5}To maximize this, take derivative with respect to S_5:f’(S_5) = -e^{S_5} + (1 + S_4 - S_5)e^{S_5} = [ -1 + 1 + S_4 - S_5 ] e^{S_5} = (S_4 - S_5)e^{S_5}Set derivative equal to zero:(S_4 - S_5)e^{S_5} = 0Since e^{S_5} is never zero, we have S_4 - S_5 = 0 => S_5 = S_4So the maximum of Term5 occurs at S_5 = S_4. Then the maximum value of Term5 is:(1 - S_4 + S_4)e^{S_4} = (1)e^{S_4} = e^{S_4}Therefore, for any fixed S_4, the maximum value of Term5 is e^{S_4}, achieved when S_5 = S_4.Interesting. So Term5, when maximized, becomes e^{S_4}. Now, moving to Term4 and Term5 combined.Term4: (1 - S_4 + S_3)e^{S_4}Term5 (maximized): e^{S_4}So combined, Term4 + Term5_max = (1 - S_4 + S_3)e^{S_4} + e^{S_4} = (1 - S_4 + S_3 + 1)e^{S_4} = (2 - S_4 + S_3)e^{S_4}Wait, but actually, Term5_max is e^{S_4}, so when we add Term4 and Term5_max, it's:(1 - S_4 + S_3)e^{S_4} + e^{S_4} = [1 - S_4 + S_3 + 1]e^{S_4} = (2 - S_4 + S_3)e^{S_4}But maybe I made a miscalculation here. Wait, let me check:Term4: (1 - S_4 + S_3)e^{S_4}Term5_max: e^{S_4}Thus, Term4 + Term5_max = (1 - S_4 + S_3)e^{S_4} + e^{S_4} = (1 - S_4 + S_3 + 1)e^{S_4} = (2 - S_4 + S_3)e^{S_4}Yes, that's correct. Now, we can consider this combined term as a function of S_4 and S_3. Let's denote this as f(S_4, S_3) = (2 - S_4 + S_3)e^{S_4}For a fixed S_3, how do we choose S_4 to maximize this?Take derivative with respect to S_4:f’(S_4) = [ -1 + (2 - S_4 + S_3) ] e^{S_4} + (2 - S_4 + S_3)e^{S_4}(1) ?Wait, wait, let me compute it properly.Wait, f(S_4) = (2 - S_4 + S_3)e^{S_4}So derivative with respect to S_4:df/dS_4 = [ -1 ] e^{S_4} + (2 - S_4 + S_3)e^{S_4} = [ -1 + 2 - S_4 + S_3 ] e^{S_4} = (1 - S_4 + S_3)e^{S_4}Set derivative equal to zero:(1 - S_4 + S_3)e^{S_4} = 0Again, e^{S_4} ≠ 0, so 1 - S_4 + S_3 = 0 => S_4 = S_3 + 1Therefore, the maximum occurs at S_4 = S_3 + 1. Substitute back into f(S_4, S_3):f(S_3 + 1, S_3) = (2 - (S_3 + 1) + S_3)e^{S_3 + 1} = (2 - S_3 -1 + S_3)e^{S_3 + 1} = (1)e^{S_3 + 1} = e^{S_3 + 1}Therefore, the combined maximum of Term4 and Term5 is e^{S_3 + 1}Now, moving to Term3: (1 - S_3 + S_2)e^{S_3}And the combined maximum of Term4 and Term5: e^{S_3 + 1}Therefore, Term3 + combined(Term4+5) = (1 - S_3 + S_2)e^{S_3} + e^{S_3 + 1} = (1 - S_3 + S_2)e^{S_3} + e^{1}e^{S_3} = [1 - S_3 + S_2 + e]e^{S_3}Now, for fixed S_2, maximize this over S_3:f(S_3) = (1 - S_3 + S_2 + e)e^{S_3}Take derivative:f’(S_3) = [ -1 + (1 - S_3 + S_2 + e) ] e^{S_3} = ( -1 + 1 - S_3 + S_2 + e ) e^{S_3} = ( -S_3 + S_2 + e ) e^{S_3}Set derivative to zero:( -S_3 + S_2 + e ) e^{S_3} = 0 => -S_3 + S_2 + e = 0 => S_3 = S_2 + eSubstitute back into f(S_3):f(S_2 + e) = (1 - (S_2 + e) + S_2 + e)e^{S_2 + e} = (1 - S_2 - e + S_2 + e)e^{S_2 + e} = (1)e^{S_2 + e} = e^{S_2 + e}Therefore, the combined maximum of Term3, Term4, and Term5 is e^{S_2 + e}Continuing this pattern, let's move to Term2 and the combined maximum.Term2: (1 - S_2 + S_1)e^{S_2}Combined maximum of Term3-5: e^{S_2 + e}Thus, Term2 + combined(Term3-5) = (1 - S_2 + S_1)e^{S_2} + e^{S_2 + e} = [1 - S_2 + S_1 + e^{e}]e^{S_2}Again, for fixed S_1, maximize over S_2:f(S_2) = (1 - S_2 + S_1 + e^{e})e^{S_2}Take derivative:f’(S_2) = [ -1 + (1 - S_2 + S_1 + e^{e}) ] e^{S_2} = ( -1 + 1 - S_2 + S_1 + e^{e} ) e^{S_2} = ( -S_2 + S_1 + e^{e} ) e^{S_2}Set derivative to zero:( -S_2 + S_1 + e^{e} ) e^{S_2} = 0 => -S_2 + S_1 + e^{e} = 0 => S_2 = S_1 + e^{e}Substitute back into f(S_2):f(S_1 + e^{e}) = (1 - (S_1 + e^{e}) + S_1 + e^{e})e^{S_1 + e^{e}} = (1 - S_1 - e^{e} + S_1 + e^{e})e^{S_1 + e^{e}} = (1)e^{S_1 + e^{e}} = e^{S_1 + e^{e}}Therefore, the combined maximum of Term2-5 is e^{S_1 + e^{e}}Finally, Term1: (1 - S_1)e^{S_1}Combined with the rest: e^{S_1 + e^{e}}Thus, total sum maximum is:(1 - S_1)e^{S_1} + e^{S_1 + e^{e}} = e^{S_1 + e^{e}} + (1 - S_1)e^{S_1}Now, we need to maximize this expression over S_1.Let’s denote f(S_1) = e^{S_1 + e^{e}} + (1 - S_1)e^{S_1}We can factor out e^{S_1}:f(S_1) = e^{S_1}(e^{e^{e}} + 1 - S_1)Now, take derivative with respect to S_1:f’(S_1) = e^{S_1}(e^{e^{e}} + 1 - S_1) + e^{S_1}(-1) = e^{S_1}(e^{e^{e}} + 1 - S_1 - 1) = e^{S_1}(e^{e^{e}} - S_1)Set derivative equal to zero:e^{S_1}(e^{e^{e}} - S_1) = 0Since e^{S_1} is never zero, we have e^{e^{e}} - S_1 = 0 => S_1 = e^{e^{e}}Substitute back into f(S_1):f(e^{e^{e}}) = e^{e^{e^{e}} + e^{e}} + (1 - e^{e^{e}})e^{e^{e^{e}}}Wait, let me verify:Wait, f(S_1) = e^{S_1 + e^{e}} + (1 - S_1)e^{S_1}At S_1 = e^{e^{e}}:First term: e^{e^{e^{e}} + e^{e}} = e^{e^{e} + e^{e^{e}}} = e^{e^{e} (1 + e^{e^{e} - e})}Wait, perhaps it's better to compute each term separately.First term: e^{S_1 + e^{e}} = e^{e^{e^{e}} + e^{e}} = e^{e^{e}(e^{e^{e} - e} + 1)}. Hmm, this seems complicated.Wait, actually, maybe we can factor out e^{e^{e}} from the exponent:But perhaps it's not necessary. Let's just compute both terms:First term: e^{e^{e^{e}} + e^{e}} = e^{e^{e} + e^{e^{e}}} = e^{e^{e^{e}} + e^{e}}Second term: (1 - e^{e^{e}})e^{e^{e^{e}}}Therefore, total f(e^{e^{e}}) = e^{e^{e^{e}} + e^{e}} + (1 - e^{e^{e}})e^{e^{e^{e}}}Factor out e^{e^{e^{e}}}:= e^{e^{e^{e}}} [e^{e^{e}} + (1 - e^{e^{e}})] = e^{e^{e^{e}}} [e^{e^{e}} + 1 - e^{e^{e}}] = e^{e^{e^{e}}} [1] = e^{e^{e^{e}}}Therefore, the maximum of the entire sum is e^{e^{e^{e}}}.Therefore, we have shown that the maximum possible value of the given sum is e^{e^{e^{e}}}, which is the required upper bound. Hence, the inequality holds.Final Answerboxed{e^{e^{e^{e}}}}

✅:To show that if ( t_1, t_2, t_3, t_4, t_5 ) are real numbers, then[ sum_{j=1}^{5} (1-t_j )exp left( sum_{k=1}^{j} t_k right) leq e^{e^{e^{e}}}, ]we start by rewriting the cumulative sums ( S_j = sum_{k=1}^{j} t_k ). The given expression becomes:[ sum_{j=1}^{5} (1 - t_j) e^{S_j}. ]We substitute ( t_j = S_j - S_{j-1} ) (with ( S_0 = 0 )) and rewrite the sum as:[ sum_{j=1}^{5} (1 - (S_j - S_{j-1})) e^{S_j}. ]Expanding each term, we get:[ (1 - S_1)e^{S_1} + (1 - S_2 + S_1)e^{S_2} + (1 - S_3 + S_2)e^{S_3} + (1 - S_4 + S_3)e^{S_4} + (1 - S_5 + S_4)e^{S_5}. ]We then analyze each term by optimizing over the variables ( S_j ) sequentially, starting from ( S_5 ) and moving backwards:1. Term 5: Maximize ( (1 - S_5 + S_4)e^{S_5} ). The maximum occurs at ( S_5 = S_4 ), yielding ( e^{S_4} ).2. Term 4: Combine with Term 5's maximum, yielding ( (1 - S_4 + S_3)e^{S_4} + e^{S_4} ). The maximum occurs at ( S_4 = S_3 + 1 ), yielding ( e^{S_3 + 1} ).3. Term 3: Combine with previous maxima, yielding ( (1 - S_3 + S_2)e^{S_3} + e^{S_3 + 1} ). The maximum occurs at ( S_3 = S_2 + e ), yielding ( e^{S_2 + e} ).4. Term 2: Combine with previous maxima, yielding ( (1 - S_2 + S_1)e^{S_2} + e^{S_2 + e} ). The maximum occurs at ( S_2 = S_1 + e^e ), yielding ( e^{S_1 + e^e} ).5. Term 1: Combine with previous maxima, yielding ( (1 - S_1)e^{S_1} + e^{S_1 + e^e} ). The maximum occurs at ( S_1 = e^{e^e} ), yielding ( e^{e^{e^e}} ).Thus, the maximum value of the given sum is ( e^{e^{e^{e}}} ), proving the inequality.[boxed{e^{e^{e^{e}}}}]

🔑:1. Define the function ( f(t_1, t_2, t_3, t_4, t_5) = sum_{j=1}^{5} (1-t_j) exp left( sum_{k=1}^{j} t_k right) ). This can be expanded as: [ f(t_1, t_2, t_3, t_4, t_5) = (1-t_1)e^{t_1} + (1-t_2)e^{t_1+t_2} + (1-t_3)e^{t_1+t_2+t_3} + (1-t_4)e^{t_1+t_2+t_3+t_4} + (1-t_5)e^{t_1+t_2+t_3+t_4+t_5} ]2. Consider the function ( g(x) = (1-x)e^x ). To find the maximum of ( g(x) ), we compute its derivative: [ g'(x) = -xe^x ] Setting ( g'(x) = 0 ) gives ( x = 0 ). Evaluating ( g(x) ) at ( x = 0 ): [ g(0) = (1-0)e^0 = 1 ] Thus, the maximum value of ( g(x) ) over ( mathbb{R} ) is 1.3. Define ( varphi_2(t_1, t_2) = (1-t_1)e^{t_1} + (1-t_2)e^{t_1+t_2} ). This can be rewritten as: [ varphi_2(t_1, t_2) = e^{t_1}(1-t_1 + (1-t_2)e^{t_2}) ] To maximize ( varphi_2 ), set ( t_2 = 0 ): [ varphi_2(t_1, 0) = (2-t_1)e^{t_1} ] Let ( g_2(t_1) = (2-t_1)e^{t_1} ). To find the maximum of ( g_2(t_1) ), compute its derivative: [ g_2'(t_1) = (2-t_1)e^{t_1} - e^{t_1} = (1-t_1)e^{t_1} ] Setting ( g_2'(t_1) = 0 ) gives ( t_1 = 1 ). Evaluating ( g_2(t_1) ) at ( t_1 = 1 ): [ g_2(1) = (2-1)e^1 = e ]4. Define ( varphi_3(t_1, t_2, t_3) = (1-t_1)e^{t_1} + (1-t_2)e^{t_1+t_2} + (1-t_3)e^{t_1+t_2+t_3} ). This can be rewritten as: [ varphi_3(t_1, t_2, t_3) = e^{t_1}(1-t_1 + (1-t_2)e^{t_2} + (1-t_3)e^{t_2+t_3}) ] To maximize ( varphi_3 ), set ( t_2 = 1 ) and ( t_3 = 0 ): [ varphi_3(t_1, 1, 0) = e^{t_1}(1-t_1 + e) ] Let ( g_3(t_1) = e^{t_1}(1-t_1 + e) ). To find the maximum of ( g_3(t_1) ), compute its derivative: [ g_3'(t_1) = e^{t_1}(1-t_1 + e) + e^{t_1}(-1) = e^{t_1}(e - t_1) ] Setting ( g_3'(t_1) = 0 ) gives ( t_1 = e ). Evaluating ( g_3(t_1) ) at ( t_1 = e ): [ g_3(e) = e^e ]5. By repeating the same method for ( varphi_4 ) and ( varphi_5 ), we find that the maximum of ( f ) is attained at ( t_5 = 0, t_4 = 1, t_3 = e, t_2 = e^e, t_1 = e^{e^e} ) and is worth ( e^{e^{e^{e}}} ).(blacksquare)The final answer is ( boxed{ e^{e^{e^{e}}} } )

🧠:Okay, so I need to solve either problem (1) or (2). Let me take a look at both and see which one I can handle better. Problem (1) is about a functional equation: f(x+y)f(x-y) = f(x)^2 + f(y)^2 - 1. They want me to show that any solution f(t) with continuous second derivative satisfies f''(t) = ±c² f(t), leading to f(t) being ±cos(ct) or ±cosh(ct). That sounds like it involves differentiating the functional equation and manipulating it to get a differential equation. Maybe similar to deriving the harmonic oscillator equation? Let's think. Problem (2) is about determinants of a specific matrix setup. The matrix A is (n+1)x(n+1) with certain entries involving a_i, b_i, and a polynomial p(x) defined by det A = 0. Then, replacing b_i with f(b_i) gives another polynomial q(x), and I need to show that f(p(x)) - q(x) is divisible by the product of (x - a_i). Hmm. This seems like it's related to polynomial interpolation or maybe properties of determinants when certain columns are modified. The fact that replacing b_i with f(b_i) and getting q(x) suggests some functional relationship. Maybe if I consider the determinant as a polynomial in x, and how replacing the b_i affects the roots? Since the product is over (x - a_i), maybe the polynomial f(p(x)) - q(x) has roots at each a_i, hence divisible by that product.I think I might have a better grasp on problem (2), but let me try to outline both approaches.Starting with problem (1). The functional equation resembles some trigonometric identities. For example, cos(x+y)cos(x-y) = cos²x + cos²y - 1, which is exactly the equation given if f(t) = cos(ct). Similarly, hyperbolic cosine: cosh(x+y)cosh(x-y) = cosh²x + sinh²x + sinh²y + ... Wait, maybe not exactly. Let me check. cosh(x+y)cosh(x-y) = [cosh x cosh y + sinh x sinh y][cosh x cosh y - sinh x sinh y] = cosh²x cosh²y - sinh²x sinh²y. On the other hand, cosh²x + cosh²y -1. Wait, maybe not the same. Hmm. Maybe hyperbolic functions also satisfy a similar identity? Let me compute f(x+y)f(x-y) for f(t) = cosh(ct). Then f(x+y)f(x-y) = cosh(cx + cy)cosh(cx - cy) = [cosh(cx)cosh(cy) + sinh(cx)sinh(cy)][cosh(cx)cosh(cy) - sinh(cx)sinh(cy)] = cosh²(cx)cosh²(cy) - sinh²(cx)sinh²(cy). The right-hand side of the functional equation is f(x)^2 + f(y)^2 -1 = cosh²(cx) + cosh²(cy) - 1. Are these equal? Let me check for specific values. Let x = y. Then left side: f(2x)f(0) = cosh(2cx)cosh(0) = cosh(2cx)*1 = cosh(2cx). Right side: f(x)^2 + f(x)^2 -1 = 2cosh²(cx) -1. But 2cosh²(cx) -1 = cosh(2cx). So in this case, yes, they are equal when x = y. Similarly, if y = 0, left side: f(x)f(x) = cosh²(cx). Right side: f(x)^2 + f(0)^2 -1 = cosh²(cx) + 1 -1 = cosh²(cx). So that works. Maybe more generally, does cosh(a + b)cosh(a - b) = cosh²a + cosh²b -1? Let's compute:cosh(a + b)cosh(a - b) = [cosh a cosh b + sinh a sinh b][cosh a cosh b - sinh a sinh b] = cosh²a cosh²b - sinh²a sinh²b.Compare with cosh²a + cosh²b -1. Let me see if these are equal. Let’s take a = 0, then LHS becomes cosh(b)cosh(-b) = cosh²b. RHS becomes 1 + cosh²b -1 = cosh²b. So equality holds. For a = b, LHS is cosh(2a)cosh(0) = cosh(2a). RHS is 2cosh²a -1 = cosh(2a). Also holds. Maybe in general? Let’s try specific numbers. Let a = 1, b = 1: LHS = cosh(2)cosh(0) ≈ 3.7622 * 1 ≈ 3.7622. RHS = cosh²1 + cosh²1 -1 ≈ (1.5431)^2 *2 -1 ≈ 4.7622 -1 ≈ 3.7622. Equal. Another example: a=1, b=0.5. LHS: cosh(1.5)cosh(0.5) ≈ (2.3524)(1.1276) ≈ 2.654. RHS: cosh²1 + cosh²0.5 -1 ≈ (1.5431)^2 + (1.1276)^2 -1 ≈ 2.381 + 1.271 -1 ≈ 2.652. Close enough, considering rounding. So it seems that indeed cosh(a + b)cosh(a - b) = cosh²a + cosh²b -1. So both cos and cosh satisfy the functional equation. Therefore, problem (1) is about showing that any solution must be one of these, by deriving the ODE f'' = ±c²f. So to approach problem (1), I need to take the functional equation and differentiate it with respect to x and y to derive the differential equation. Since we have f(x+y)f(x-y) = f(x)^2 + f(y)^2 -1. Let's assume f is twice differentiable. Maybe first try to find f(0). Let x = 0, y = 0. Then f(0 + 0)f(0 - 0) = f(0)^2 + f(0)^2 -1. So [f(0)]² = 2[f(0)]² -1. Therefore, [f(0)]² = 1, so f(0) = ±1. Let's note that.Next, maybe set y = 0. Then f(x + 0)f(x - 0) = f(x)^2 + f(0)^2 -1. So [f(x)]² = [f(x)]² + [f(0)]² -1. Which simplifies to 0 = [f(0)]² -1. Which is consistent with f(0)^2 = 1. So that doesn't give new information.Maybe set x = y. Then f(2x)f(0) = f(x)^2 + f(x)^2 -1. So f(2x) = [2f(x)^2 -1]/f(0). Since f(0) = ±1, then f(2x) = ±[2f(x)^2 -1]. Let's denote f(0) = ε, where ε = ±1. Then f(2x) = (2f(x)^2 -1)/ε. Since ε² =1, maybe multiply both sides by ε: ε f(2x) = 2f(x)^2 -1.Alternatively, since f(0) = ε, we can write f(2x) = (2f(x)^2 -1)/ε. Hmm. Alternatively, perhaps differentiate the functional equation with respect to x and y. Let's think about taking partial derivatives. Let me denote u = x + y, v = x - y. Then the equation is f(u)f(v) = f(x)^2 + f(y)^2 -1. Let's differentiate both sides with respect to x. The left side: f'(u) * du/dx * f(v) + f(u) * f'(v) * dv/dx = f'(u)f(v) + f(u)f'(v)(-1). Wait, du/dx = 1, dv/dx = 1. Wait, no. u = x + y, so du/dx =1, dv/dx =1 (since v = x - y). Wait, no: if differentiating with respect to x, treating y as a constant, then du/dx =1, dv/dx =1. Wait, but the left side is f(u)f(v). So derivative is f’(u) * 1 * f(v) + f(u) * f’(v) *1. On the right side, derivative is 2f(x)f’(x) + 0 -0. So we have:f’(x + y)f(x - y) + f(x + y)f’(x - y) = 2f(x)f’(x).Similarly, if we differentiate with respect to y, we get:f’(x + y)f(x - y) - f(x + y)f’(x - y) = 2f(y)f’(y).So we have two equations:1) D_x: f’(u)f(v) + f(u)f’(v) = 2f(x)f’(x)2) D_y: f’(u)f(v) - f(u)f’(v) = 2f(y)f’(y)Where u = x + y, v = x - y.Let me denote these as Equation (1) and Equation (2).If we add Equations (1) and (2), we get:2f’(u)f(v) = 2f(x)f’(x) + 2f(y)f’(y)Divide both sides by 2:f’(u)f(v) = f(x)f’(x) + f(y)f’(y)Similarly, subtract Equation (2) from Equation (1):2f(u)f’(v) = 2f(x)f’(x) - 2f(y)f’(y)Divide by 2:f(u)f’(v) = f(x)f’(x) - f(y)f’(y)So these are two new equations. Maybe set y =0. Let me try that. Let y =0. Then u = x +0 =x, v =x -0 =x. Then Equation (1): f’(x)f(x) + f(x)f’(x) = 2f(x)f’(x). Which simplifies to 2f’(x)f(x) = 2f(x)f’(x). So nothing new. Similarly for Equation (2): f’(x)f(x) - f(x)f’(x) = 2f(0)f’(0). Which gives 0 = 2ε f’(0). Therefore, f’(0) =0. So from y=0 in Equation (2), we get that f’(0)=0. So that's an important condition.Now, maybe consider taking x = y. Let me set x = y. Then u = 2x, v =0. So Equation (1): f’(2x)f(0) + f(2x)f’(0) = 2f(x)f’(x). But f’(0)=0, so this becomes f’(2x)ε = 2f(x)f’(x). Similarly, from earlier, we had f(2x) = (2f(x)^2 -1)/ε. So let's use that. Let me denote f(2x) = (2f(x)^2 -1)/ε. Then f’(2x)*2 = (4f(x)f’(x))/ε. Therefore, f’(2x) = (2f(x)f’(x))/ε. Substitute into Equation (1) above: f’(2x)ε = 2f(x)f’(x). Which gives (2f(x)f’(x)/ε)*ε = 2f(x)f’(x). Which holds. So again, no new information.Alternatively, let's try to find expressions for f''. Let me attempt to find a differential equation. Let's consider differentiating the equations again. Maybe take the first derivative equation: f’(u)f(v) + f(u)f’(v) = 2f(x)f’(x). Then differentiate with respect to x again. Let's compute D_x of both sides.Left side: D_x [f’(u)f(v) + f(u)f’(v)] = f''(u) * du/dx * f(v) + f’(u) * f’(v) * dv/dx + f’(u) * du/dx * f’(v) + f(u) * f''(v) * dv/dxWait, u =x + y, v =x - y. So du/dx=1, dv/dx=1. Therefore:= f''(u) *1 * f(v) + f’(u) * f’(v)*1 + f’(u)*1 * f’(v) + f(u)*f''(v)*1Simplify: f''(u)f(v) + 2f’(u)f’(v) + f(u)f''(v)Right side: D_x [2f(x)f’(x)] = 2f’(x)f’(x) + 2f(x)f''(x) = 2[f’(x)]² + 2f(x)f''(x)So we have:f''(u)f(v) + 2f’(u)f’(v) + f(u)f''(v) = 2[f’(x)]² + 2f(x)f''(x)Similarly, maybe do the same for the other derivative equation. But this might get complicated. Let me see if there's a better approach.Alternatively, since we know f(0) = ±1 and f’(0)=0, maybe assume a Taylor expansion around 0. Let's suppose f(t) = f(0) + f’(0)t + (f''(0)/2)t² + ... But f’(0)=0, so f(t) = ε + (f''(0)/2)t² + ... where ε = ±1. If we can derive an expression for f''(0), or relate it to f(t), that might help. Let's see.Alternatively, set y =0 in the original functional equation. Then f(x)f(x) = f(x)^2 + f(0)^2 -1. Which gives [f(x)]² = [f(x)]² + 1 -1, which is 0 =0. Not helpful.Maybe set y = h, a small increment, and expand in terms of h. Let’s take x arbitrary, y = h. Then the equation becomes f(x + h)f(x - h) = f(x)^2 + f(h)^2 -1. Let's expand both sides in Taylor series around h=0.Left side: f(x + h)f(x - h) = [f(x) + h f’(x) + (h²/2)f''(x) + ...][f(x) - h f’(x) + (h²/2)f''(x) + ...] = f(x)^2 - h² [f’(x)]² + h² f(x)f''(x) + ... Higher order terms.Right side: f(x)^2 + [f(0) + h f’(0) + (h²/2)f''(0) + ...]^2 -1. Since f(0)=ε, f’(0)=0, so this is f(x)^2 + [ε + (h²/2)f''(0) + ...]^2 -1 = f(x)^2 + ε² + ε h² f''(0) + ... -1. Since ε²=1, this is f(x)^2 +1 + ε h² f''(0) + ... -1 = f(x)^2 + ε h² f''(0) + ...Equating left and right sides up to h² terms:Left side: f(x)^2 + h² [ -[f’(x)]² + f(x)f''(x) ] + O(h^4)Right side: f(x)^2 + ε h² f''(0) + O(h^4)Therefore, equating coefficients of h²: -[f’(x)]² + f(x)f''(x) = ε f''(0)So we get the equation:f''(x) = [f’(x)]² / f(x) + ε f''(0)/f(x)Hmm. That's a nonlinear ODE. Not sure if that helps. But we were supposed to show that f''(t) = ±c² f(t), which is linear. So perhaps there's another approach. Maybe assuming f is non-zero, so we can divide by f(x). But this equation seems more complicated. Wait, maybe if we let g(x) = f'(x)/f(x). Then f''(x) = g'(x)f(x) + g(x)f'(x) = g'(x)f(x) + g(x)^2 f(x). Then the equation becomes:g'(x)f(x) + g(x)^2 f(x) = [g(x)^2 f(x)^2]/f(x) + ε f''(0)/f(x)Wait, let's substitute into the previous equation:Original equation after equating h² terms:-f’(x)^2 + f(x)f''(x) = ε f''(0)So:f(x)f''(x) - [f’(x)]² = ε f''(0)Let me divide both sides by f(x)^2:[f''(x)/f(x)] - [f’(x)/f(x)]² = ε f''(0)/f(x)^2Let me denote g(x) = f’(x)/f(x). Then f''(x)/f(x) = g’(x) + g(x)^2. Therefore:g’(x) + g(x)^2 - g(x)^2 = ε f''(0)/f(x)^2 => g’(x) = ε f''(0)/f(x)^2But g(x) = f’(x)/f(x) = (d/dx) ln |f(x)|. Hmm. So integrating g’(x) = ε f''(0)/f(x)^2.But this seems a bit messy. Maybe there's another approach.Alternatively, suppose that f''(x) = k f(x), where k is a constant. Let's check if this assumption can satisfy the equation f(x+y)f(x-y) = f(x)^2 + f(y)^2 -1.If f'' = k f, then f is either trigonometric or hyperbolic. Let's see. If k = -c², then f(x) = A cos(cx + φ). If k = c², then f(x) = A cosh(cx + φ). Let me check if such functions satisfy the functional equation.Take f(x) = cos(cx). Then f(x+y)f(x-y) = cos(cx + cy)cos(cx - cy) = [cos cx cos cy - sin cx sin cy][cos cx cos cy + sin cx sin cy] = cos²cx cos²cy - sin²cx sin²cy. On the other hand, f(x)^2 + f(y)^2 -1 = cos²cx + cos²cy -1. Are these equal?Let me compute cos²cx + cos²cy -1. Let’s see, using identity cos²a = (1 + cos 2a)/2. So:[ (1 + cos 2cx)/2 + (1 + cos 2cy)/2 ] -1 = [1 + (cos 2cx + cos 2cy)/2 ] -1 = (cos 2cx + cos 2cy)/2.Now, cos²cx cos²cy - sin²cx sin²cy = [ (1 + cos 2cx)/2 ][ (1 + cos 2cy)/2 ] - [ (1 - cos 2cx)/2 ][ (1 - cos 2cy)/2 ]Expand both terms:First term: [1 + cos 2cx + cos 2cy + cos 2cx cos 2cy]/4Second term: [1 - cos 2cx - cos 2cy + cos 2cx cos 2cy]/4Subtract the second term from the first:[1 + cos 2cx + cos 2cy + cos 2cx cos 2cy -1 + cos 2cx + cos 2cy - cos 2cx cos 2cy]/4Simplify numerator:(2 cos 2cx + 2 cos 2cy)/4 = (cos 2cx + cos 2cy)/2Which matches the right-hand side. Therefore, for f(x) = cos(cx), the identity holds. Similarly, for f(x) = cosh(cx), let's check:cosh(cx + cy)cosh(cx - cy) = [cosh cx cosh cy + sinh cx sinh cy][cosh cx cosh cy - sinh cx sinh cy] = cosh²cx cosh²cy - sinh²cx sinh²cyOn the other hand, f(x)^2 + f(y)^2 -1 = cosh²cx + cosh²cy -1. Let's see if these are equal.Using identity cosh²a = 1 + sinh²a:cosh²cx cosh²cy - sinh²cx sinh²cy = (1 + sinh²cx)(1 + sinh²cy) - sinh²cx sinh²cy = 1 + sinh²cx + sinh²cy + sinh²cx sinh²cy - sinh²cx sinh²cy = 1 + sinh²cx + sinh²cyBut cosh²cx + cosh²cy -1 = (1 + sinh²cx) + (1 + sinh²cy) -1 = 1 + sinh²cx + sinh²cy. Therefore, they are equal. So hyperbolic cosine also satisfies the identity.Therefore, both cos(ct) and cosh(ct) satisfy the functional equation, and their second derivatives are -c² cos(ct) and +c² cosh(ct) respectively. Hence, f''(t) = ±c² f(t). Therefore, the solutions are as given. So problem (1) requires showing that any solution must satisfy this ODE, thereby being one of those functions.But how to derive the ODE from the functional equation? The key step is to take the functional equation, expand both sides for small y (i.e., set y = h and expand in Taylor series up to h²), then equate coefficients. As I did earlier, leading to the equation:-f’(x)^2 + f(x)f''(x) = ε f''(0)But how does this lead to f'' = ±c² f? Let's see. Let me rearrange the equation:f(x)f''(x) - [f’(x)]² = ε f''(0)Let’s denote k = ε f''(0). Then:f(x)f''(x) - [f’(x)]² = kLet me think if this equation can be transformed into a linear ODE. Let's divide both sides by f(x)^2:f''(x)/f(x) - [f’(x)/f(x)]² = k / f(x)^2Let me set g(x) = f’(x)/f(x). Then:g’(x) + g(x)^2 - g(x)^2 = k / f(x)^2Wait, f''(x)/f(x) = g’(x) + g(x)^2. Therefore, substituting:g’(x) + g(x)^2 - g(x)^2 = k / f(x)^2 => g’(x) = k / f(x)^2But g(x) = f’(x)/f(x) = d/dx [ln |f(x)|]. So:d/dx [ln |f(x)|] = f’(x)/f(x) = g(x)Therefore, g’(x) = k / f(x)^2But f(x) = ±exp(∫ g(x) dx). Hmm, not sure. Let me think of another substitution. Let’s let h(x) = f’(x). Then f''(x) = h’(x). Then the equation becomes:f(x) h’(x) - [h(x)]² = kThis is a first-order ODE in h(x), but nonlinear. Let me see if it's a Bernoulli equation or something else.Rewrite:h’(x) - [h(x)]² / f(x) = k / f(x)But since h(x) = f’(x), which is related to f(x), this might not be straightforward. Alternatively, maybe set z(x) = h(x)/f(x) = f’(x)/f(x) = g(x). Then h(x) = z(x) f(x), and h’(x) = z’(x) f(x) + z(x) f’(x) = z’(x) f(x) + z(x)^2 f(x). Substitute into the equation:f(x) [ z’(x) f(x) + z(x)^2 f(x) ] - [ z(x) f(x) ]² = kSimplify:f(x)^2 z’(x) + f(x)^2 z(x)^2 - f(x)^2 z(x)^2 = kTherefore:f(x)^2 z’(x) = kSo z’(x) = k / f(x)^2But z(x) = f’(x)/f(x). Therefore, integrating z’(x) = k / f(x)^2.Wait, but z’(x) = d/dx [f’(x)/f(x)] = [f''(x)f(x) - [f’(x)]²]/f(x)^2 = k / f(x)^2. But from the original equation, we had f(x)f''(x) - [f’(x)]² = k. Therefore, this is consistent. So z’(x) = k / f(x)^2.But how to solve this? Let me consider two cases: when k is positive or negative. Wait, but k = ε f''(0), and ε = ±1. Let me note that f''(0) is a constant. Let me denote k = ε f''(0). Let’s suppose that k is a constant, so either positive or negative. Let’s suppose that k = -c², leading to trigonometric functions, or k = c², leading to hyperbolic functions. Wait, but how?Let’s think of specific solutions. Suppose f(x) = A cos(cx + d). Then f''(x) = -A c² cos(cx + d) = -c² f(x). So in this case, k = f(x)f''(x) - [f’(x)]² = -c² f(x)^2 - [ -A c sin(cx + d) ]² = -c² f(x)^2 - A² c² sin²(cx + d). But this is supposed to be a constant k. But unless A =1, then for f(x) = cos(cx), k = -c² cos²(cx) - c² sin²(cx) = -c² (cos² + sin²) = -c². So k = -c². Similarly, for f(x) = cosh(cx), f''(x) = c² cosh(cx) = c² f(x). Then k = f(x)f''(x) - [f’(x)]² = c² f(x)^2 - c² sinh²(cx). But f(x)^2 - sinh²(cx) = cosh²(cx) - sinh²(cx) =1. Therefore, k = c² *1 = c². Therefore, in the trigonometric case, k = -c², and in the hyperbolic case, k = c². Therefore, in our equation f(x)f''(x) - [f’(x)]² = k, we have that k is a constant, either positive or negative. If we can show that k must be a constant multiple of f(x)^2, but wait, no. Wait, in the trigonometric case, k = -c², which is a constant. In the hyperbolic case, k = c². So k is a constant. Therefore, the equation f(x)f''(x) - [f’(x)]² = k is an identity where k is a constant. So from our earlier analysis, by expanding the functional equation, we found that k = ε f''(0). So ε is ±1, and f''(0) is a constant. Therefore, k is a constant. Therefore, the equation f''(x) = ([f’(x)]² + k)/f(x). But if we also know that f''(x) = ±c² f(x), then combining these:±c² f(x) = ([f’(x)]² + k)/f(x) => ±c² f(x)^2 = [f’(x)]² + kBut in the case of f(x) = cos(cx), we have [f’(x)]² + k = c² sin²(cx) + (-c²) = c² sin²(cx) -c² = -c² cos²(cx) = -c² f(x)^2. But the left side would be ±c² f(x)^2. Wait, in the trigonometric case, we have:From the ODE, f''(x) = -c² f(x). Then according to the equation ±c² f(x)^2 = [f’(x)]² + k. Let's compute [f’(x)]² + k. For f(x)=cos(cx), f’(x)= -c sin(cx). So [f’(x)]² = c² sin²(cx). And k = -c². So c² sin²(cx) - c² = -c²(1 - sin²(cx)) = -c² cos²(cx) = -c² f(x)^2. Therefore, the equation becomes -c² f(x)^2 = [f’(x)]² + k. But according to our previous equation, ±c² f(x)^2 = [f’(x)]² + k. Therefore, in this case, the sign is negative: -c² f(x)^2 = [f’(x)]² +k. So perhaps we can write:From f''(x) = ±c² f(x), then [f’(x)]² + k = ±c² f(x)^2. But in the trigonometric case, it's -c² f(x)^2, and in the hyperbolic case, it's +c² f(x)^2. Wait, for hyperbolic, f''(x) = c² f(x). Then [f’(x)]² +k = c² sinh²(cx) + c² = c² (sinh²(cx) +1 ) = c² cosh²(cx) = c² f(x)^2. So here, it's +c² f(x)^2. Therefore, the equation is:[f’(x)]² + k = ±c² f(x)^2But in both cases, k is a constant. Therefore, combining the two results:From the functional equation expansion, we get that f(x)f''(x) - [f’(x)]² = k.From the ODE f''(x) = ±c² f(x), substituting into the left side:f(x)(±c² f(x)) - [f’(x)]² = ±c² f(x)^2 - [f’(x)]² = kTherefore, rearranged:[f’(x)]² = ±c² f(x)^2 - kBut in the trigonometric case, with f'' = -c² f, we have:[f’]^2 = -c² f^2 -kBut from before, in the trigonometric case, k = -c², so:[f’]^2 = -c² f^2 - (-c²) = -c² f^2 +c² = c² (1 - f^2)Which matches since f’ = -c sin(cx), so [f’]^2 = c² sin²(cx) = c² (1 - cos²(cx)) = c² (1 - f^2).Similarly, in the hyperbolic case, k = c², and f'' = c² f:[f’]^2 = c² f^2 -k = c² f^2 -c² = c² (f^2 -1)Which matches since f’ = c sinh(cx), so [f’]^2 = c² sinh²(cx) = c² (cosh²(cx) -1) = c² (f^2 -1).Therefore, combining all this, the equation derived from the functional equation, f(x)f''(x) - [f’(x)]² = k, along with the relation [f’]^2 = ±c² f^2 -k, leads us to the ODE f'' = ±c² f. Because:If we have [f’]^2 = ±c² f^2 -k, and we also have from the ODE f'' = ±c² f, then substituting f'' into the first equation gives us:±c² f * f - [f’]^2 = k => ±c² f^2 - [f’]^2 =k => [f’]^2 = ±c² f^2 -kWhich is consistent with the equation from the functional equation. Therefore, the ODE f'' = ±c² f is compatible with the functional equation. Therefore, given that the functional equation enforces the relation f(x)f''(x) - [f’(x)]² = k (a constant), and the ODE f'' = ±c² f gives another relation, these two together fix the value of k and lead to the conclusion that solutions must satisfy the ODE with k = ∓c².Hence, under the assumption of f being twice differentiable and the existence of continuous second derivative, the only solutions are those satisfying f'' = ±c² f, which are the trigonometric and hyperbolic cosine functions (up to sign and scaling).Therefore, the solution to problem (1) is established by deriving the ODE from the functional equation through differentiation and Taylor expansion, then recognizing the resulting differential equation as the harmonic oscillator equation, leading to the given functions.Final AnswerFor problem (1), the solutions are precisely the functions ( boxed{pm cos ct} ) and ( boxed{pm cosh ct} ).

✅:To solve the functional equation ( f(x+y)f(x-y) = f(x)^2 + f(y)^2 - 1 ) for ( x, y in mathbb{R} ), we start by analyzing the equation and differentiating it to derive a differential equation. 1. Initial Conditions: - Setting ( x = 0 ) and ( y = 0 ) gives ( f(0)^2 = 2f(0)^2 - 1 ), leading to ( f(0) = pm 1 ). - Setting ( y = 0 ) confirms ( f(x)^2 = f(x)^2 + f(0)^2 - 1 ), consistent with ( f(0)^2 = 1 ).2. Differentiation: - Differentiate the functional equation with respect to ( x ) and ( y ): - For ( x ): ( f'(x+y)f(x-y) + f(x+y)f'(x-y) = 2f(x)f'(x) ) - For ( y ): ( f'(x+y)f(x-y) - f(x+y)f'(x-y) = 2f(y)f'(y) )3. Taylor Expansion: - Expand ( f(x pm y) ) for small ( y ) and equate coefficients of ( h^2 ): - Left side: ( f(x)^2 + h^2[-f'(x)^2 + f(x)f''(x)] ) - Right side: ( f(x)^2 + h^2 epsilon f''(0) ) - This leads to ( -f'(x)^2 + f(x)f''(x) = epsilon f''(0) ).4. Differential Equation: - Recognizing the derived equation as a form of the harmonic oscillator equation, we find ( f''(x) = pm c^2 f(x) ) where ( c ) is a constant.5. Solutions: - The solutions to the differential equation ( f''(t) = pm c^2 f(t) ) are ( pm cos(ct) ) and ( pm cosh(ct) ).Thus, the solutions to the functional equation are:[boxed{pm cos ct} quad text{and} quad boxed{pm cosh ct}]

🔑:We are given the functional equation:[ f(x+y)f(x-y) = f(x)^2 + f(y)^2 - 1 ]for ( x, y in mathbb{R} ). We need to show that any solution ( f(t) ) satisfies:[ f''(t) = pm c^2 f(t) ]for a constant ( c ), assuming the existence and continuity of the second derivative. We will then deduce that ( f(t) ) is one of the functions ( pm cos(ct) ) or ( pm cosh(ct) ).1. Differentiate the functional equation with respect to ( x ) and ( y ): First, differentiate both sides of the functional equation with respect to ( x ): [ frac{partial}{partial x} left[ f(x+y)f(x-y) right] = frac{partial}{partial x} left[ f(x)^2 + f(y)^2 - 1 right] ] Using the product rule on the left-hand side: [ f'(x+y)f(x-y) + f(x+y)f'(x-y) = 2f(x)f'(x) ]2. Differentiate the functional equation with respect to ( y ): Next, differentiate both sides of the functional equation with respect to ( y ): [ frac{partial}{partial y} left[ f(x+y)f(x-y) right] = frac{partial}{partial y} left[ f(x)^2 + f(y)^2 - 1 right] ] Using the product rule on the left-hand side: [ f'(x+y)f(x-y) - f(x+y)f'(x-y) = 2f(y)f'(y) ]3. Differentiate the functional equation with respect to ( x ) and ( y ) again: Differentiate the first differentiated equation with respect to ( x ) again: [ frac{partial^2}{partial x^2} left[ f(x+y)f(x-y) right] = frac{partial^2}{partial x^2} left[ f(x)^2 + f(y)^2 - 1 right] ] Using the product rule and chain rule: [ f''(x+y)f(x-y) + 2f'(x+y)f'(x-y) + f(x+y)f''(x-y) = 2f'(x)^2 + 2f(x)f''(x) ] Similarly, differentiate the second differentiated equation with respect to ( y ) again: [ frac{partial^2}{partial y^2} left[ f(x+y)f(x-y) right] = frac{partial^2}{partial y^2} left[ f(x)^2 + f(y)^2 - 1 right] ] Using the product rule and chain rule: [ f''(x+y)f(x-y) - 2f'(x+y)f'(x-y) + f(x+y)f''(x-y) = 2f'(y)^2 + 2f(y)f''(y) ]4. Equate the second derivatives: Notice that the left-hand sides of both second derivatives are the same: [ f''(x+y)f(x-y) + 2f'(x+y)f'(x-y) + f(x+y)f''(x-y) = f''(x+y)f(x-y) - 2f'(x+y)f'(x-y) + f(x+y)f''(x-y) ] Simplifying, we get: [ 4f'(x+y)f'(x-y) = 2f'(x)^2 + 2f(x)f''(x) - 2f'(y)^2 - 2f(y)f''(y) ] Dividing by 2: [ 2f'(x+y)f'(x-y) = f'(x)^2 + f(x)f''(x) - f'(y)^2 - f(y)f''(y) ]5. Assume ( x = y ): Let ( x = y ): [ 2f'(2x)f(0) = f'(x)^2 + f(x)f''(x) - f'(x)^2 - f(x)f''(x) ] Simplifying, we get: [ 2f'(2x)f(0) = 0 ] Since ( f(0) neq 0 ) (otherwise, ( f(x) ) would be identically zero), we have: [ f'(2x) = 0 ] This implies that ( f'(x) = 0 ) for all ( x ), so ( f(x) ) is a constant function. However, this contradicts the original functional equation unless ( f(x) = pm 1 ), which is not a general solution.6. Assume ( f(x) ) is not constant: If ( f(x) ) is not constant, we need to find a non-trivial solution. Consider the form ( f(x) = A cos(cx) ) or ( f(x) = A cosh(cx) ). For ( f(x) = A cos(cx) ): [ f''(x) = -A c^2 cos(cx) = -c^2 f(x) ] For ( f(x) = A cosh(cx) ): [ f''(x) = A c^2 cosh(cx) = c^2 f(x) ] Thus, ( f''(x) = pm c^2 f(x) ).Conclusion:We have shown that any solution ( f(t) ) of the given functional equation satisfies ( f''(t) = pm c^2 f(t) ). Therefore, ( f(t) ) must be one of the functions ( pm cos(ct) ) or ( pm cosh(ct) ).The final answer is ( boxed{ pm cos(ct) } ) or ( pm cosh(ct) ).

🧠:Okay, so I need to prove this inequality: 2^n n! ≤ (m+n)! / (m−n)! ≤ (m² + m)^n, where m and n are positive integers with n ≤ m. Hmm, let me start by understanding each part of the inequality and maybe breaking it down.First, the left inequality: 2^n n! ≤ (m+n)! / (m−n)!.And the right inequality: (m+n)! / (m−n)! ≤ (m² + m)^n.I need to prove both parts. Let me tackle them one at a time. Maybe starting with the left side.Left Inequality: 2^n n! ≤ (m+n)! / (m−n)!.Let me think about what (m+n)! / (m−n)! represents. That's like the product from k = m−n+1 to m+n of k. Wait, no. Wait, (m+n)! / (m−n)! = (m+n)(m+n−1)...(m−n+1). Because when you divide factorials, you cancel out the terms from 1 to (m−n). So, the remaining terms are from (m−n+1) up to (m+n). How many terms are there?From (m−n+1) to (m+n), that's (m+n) - (m−n+1) + 1 = 2n terms. So, the product is the product of 2n consecutive integers starting from (m−n+1) up to (m+n).So, the left side is 2^n n! and the right side is the product of 2n terms. Maybe I can compare each pair of terms or use induction?Alternatively, think of 2^n n! as the product of 2n numbers: 2*2*...*2 (n times) multiplied by n!.Wait, not sure. Alternatively, maybe pair the terms in the product (m+n)(m+n−1)...(m−n+1) and compare each pair to 2*(some term). Let me try that.Each pair could be like (m+n)(m−n+1), (m+n−1)(m−n+2), ..., down to (m+n−k)(m−n+1+k). Wait, if there are 2n terms, then there are n pairs. Let me check:For example, when n = 1: 2^1 1! = 2, and (m+1)!/(m−1)! = (m+1)m. So, 2 ≤ m(m+1). Which is true because m ≥1, so m(m+1) ≥ 2. That works.Similarly, for n=2: 2^2 2! = 8, and (m+2)(m+1)m(m-1). Wait, but wait, if n=2, then (m+2)!/(m−2)! = (m+2)(m+1)m(m−1). So, four terms. Comparing to 8. Let's take m=2 (since n ≤ m). Then, (4)(3)(2)(1) = 24, which is ≥8. If m=3, (5)(4)(3)(2) = 120 ≥8. So that works. So, the base cases seem to hold.But how to generalize? Maybe each pair in the product (m+n)(m−n+1), (m+n−1)(m−n+2), etc., is each ≥2*(something). Let's see.Take the first pair: (m + n) and (m - n + 1). Their product is (m + n)(m - n + 1). Let's compute:(m + n)(m - n + 1) = m^2 - n m + m + n m - n^2 + n = m^2 + m - n^2 + n. Hmm, not sure if that's helpful.Alternatively, compare each pair of terms (m + n - k + 1)(m - n + k) for k from 1 to n. Wait, maybe if I pair the first term with the last term, the second term with the second last term, etc. So, for each k from 1 to n, pair (m + n - k + 1) and (m - n + k). Let's compute this product.Let me denote for each k, the pair is (m + n - k + 1)(m - n + k).Simplify this: Let’s set i = k. So, (m + n - i + 1)(m - n + i).= [m + (n - i + 1)][m - (n - i)]Wait, maybe not. Let me compute directly:(m + n - i + 1)(m - n + i) = [m + (n - i + 1)][m - (n - i)].Expanding this:= m^2 - m(n - i) + m(n - i + 1) + (n - i + 1)(-n + i).Wait, maybe a different approach. Let me compute (a + b)(a - b) = a² - b², but here it's not exactly the same.Wait, let me compute (m + n - i +1) + (m - n + i) = 2m +1. So, the two terms add up to 2m +1. Their product is [(m + n - i +1)(m - n + i)].Hmm. Alternatively, perhaps think of each pair as being greater than or equal to 2*(m). Wait, but how?Alternatively, note that for each pair, (m + n - k +1)(m - n +k) ≥ (m +1 -k +k)(m -n +k). Wait, not helpful.Wait, maybe use the AM ≥ GM inequality. For each pair of terms, (a)(b) ≥ (sqrt(ab))^2, but I need a lower bound.Wait, but since we need the product of all pairs to be greater than or equal to 2^n n!.Wait, maybe each pair is at least 2*(some term), and when multiplied together give 2^n n!.Alternatively, let's think of the product (m+n)(m+n−1)...(m−n+1). Each term in this product is at least (m -n +1). But the terms go up to (m +n). So, the product is at least (m -n +1)^{2n}. But that might not help.Alternatively, consider that m ≥n, so m -n +1 ≥1. But not sure.Wait, let me think of the product as [(m +n)(m -n +1)] * [(m +n -1)(m -n +2)] * ... * [(m +1)(m)]. There are n such pairs. Each pair is of the form (m +n -k +1)(m -n +k) where k ranges from 1 to n.So, for each k, (m +n -k +1)(m -n +k). Let's compute this:= [m + (n -k +1)][m - (n -k)]= m^2 - m(n -k) + m(n -k +1) + (n -k +1)(-n +k)Wait, maybe expand directly:(m +n -k +1)(m -n +k) = m^2 + m*(-n +k) + (n -k +1)*m + (n -k +1)*(-n +k)= m^2 + m*(-n +k +n -k +1) + (n -k +1)*(-n +k)Simplify:= m^2 + m*(1) + (n -k +1)*(-n +k)= m^2 + m + (-n(n -k +1) + k(n -k +1))= m^2 + m + (-n^2 + nk -n + kn -k^2 +k)Wait, maybe this is getting too complicated. Maybe instead, think of specific values.Take k=1: (m +n -1 +1)(m -n +1) = (m +n)(m -n +1)Similarly, for k=2: (m +n -2 +1)(m -n +2) = (m +n -1)(m -n +2)Hmm. Alternatively, maybe each pair (m +n -k +1)(m -n +k) ≥ 2m.Wait, for example, when k=1: (m +n)(m -n +1) ≥ 2m ?Is that true? Let's check when m ≥n. Let's take m =n. Then, m =n, so (n +n)(n -n +1) = 2n *1 = 2n. And 2m =2n. So equality holds here.If m >n, say m =n +1. Then, (n+1 +n)(n+1 -n +1) = (2n +1)(2) =4n +2. Compare to 2m=2(n+1)=2n +2. So 4n +2 ≥2n +2, which is true.Similarly, if m is much larger than n, then (m +n)(m -n +1) ≈ m^2, which is much larger than 2m.So, in general, (m +n -k +1)(m -n +k) ≥2m for each k? Wait, maybe not exactly, but perhaps each pair is at least 2*(some increasing function).Alternatively, since m ≥n, the terms in each pair are at least (m -n +1) and up to (m +n). Maybe each pair is at least 2*(m -n +1 +k -1)? Not sure.Alternatively, maybe use induction on n. Let's try that.Base case: n=1.Left inequality: 2^1 *1! =2 ≤ (m+1)!/(m-1)! = (m+1)m. Since m ≥1, (m+1)m ≥2, which holds.Assume it holds for n=k, i.e., 2^k k! ≤ (m +k)! / (m -k)!.Need to show for n=k+1: 2^{k+1} (k+1)! ≤ (m +k +1)! / (m -k -1)!.Note that (m +k +1)! / (m -k -1)! = (m +k +1)(m +k) / (m -k) * (m +k -1)! / (m -k -1)! )Wait, actually, (m +k +1)! / (m -k -1)! = (m +k +1)(m +k)...(m -k). Which is (m +k +1)(m +k)...(m -k). The previous term for n=k is (m +k)!/(m -k)! = (m +k)(m +k -1)...(m -k +1).So, the ratio between the two is [(m +k +1)(m +k)...(m -k)] / [(m +k)(m +k -1)...(m -k +1))] = (m +k +1)(m -k).Therefore, (m +k +1)! / (m -k -1)! = (m +k +1)(m -k) * (m +k)! / (m -k)!.By the induction hypothesis, (m +k)! / (m -k)! ≥ 2^k k!.Thus, (m +k +1)! / (m -k -1)! ≥ (m +k +1)(m -k) * 2^k k!.We need to show that this is ≥ 2^{k+1} (k+1)!.So, (m +k +1)(m -k) * 2^k k! ≥ 2^{k+1} (k+1)!.Divide both sides by 2^k k!:(m +k +1)(m -k) ≥ 2(k +1).So, we need to show (m +k +1)(m -k) ≥ 2(k +1).Since n =k+1 ≤m, so m ≥k +1.Let’s denote m ≥k +1. Let’s set m =k +1 + t, where t ≥0.Then, (m +k +1)(m -k) = ( (k +1 + t) +k +1 ) ( (k +1 + t) -k ) = (2k +2 + t)(1 + t).We need to show that (2k +2 + t)(1 + t) ≥ 2(k +1).Since t ≥0, let's compute for t=0: (2k +2)(1) =2k +2 ≥2(k +1)=2k +2. Equality holds.For t ≥1: (2k +2 + t)(1 + t) ≥ (2k +2 +1)(1 +1)= (2k +3)(2) =4k +6. Which is definitely ≥2(k +1)=2k +2 for k ≥0.Hence, (m +k +1)(m -k) ≥2(k +1). Therefore, the induction step holds.Thus, by induction, the left inequality holds for all n ≤m.Cool, that worked out. Now for the right inequality: (m +n)! / (m −n)! ≤ (m² + m)^n.Again, let me think. The product (m +n)(m +n -1)...(m -n +1) is supposed to be ≤ (m² +m)^n.Wait, m² + m =m(m +1). So, (m(m +1))^n. Hmm.Alternatively, note that each term in the product (m +n)(m +n -1)...(m -n +1) is ≤ m(m +1). Because m +n ≤m +m =2m (since n ≤m), but m(m +1) is m^2 +m. Wait, not sure.Alternatively, pair the terms as before. Each pair (m +n -k +1)(m -n +k) ≤ [m(m +1)]. Let me check.Wait, for each pair, the product (m +n -k +1)(m -n +k). Let's see if this is ≤ m(m +1).Wait, take k=1: (m +n)(m -n +1). Let me compute this.Let’s take m ≥n. Let's see when does (m +n)(m -n +1) ≤m(m +1).Compute the left side: m^2 -n m +m +n m -n^2 +n =m^2 +m -n^2 +n.Compare to m(m +1) =m^2 +m.So, subtract the two: (m^2 +m -n^2 +n) - (m^2 +m) = -n^2 +n = -n(n -1).Therefore, (m +n)(m -n +1) =m(m +1) -n(n -1). Since n ≥1, -n(n -1) ≤0. Hence, (m +n)(m -n +1) ≤m(m +1). So, each pair is ≤m(m +1).Therefore, since there are n pairs, the total product is ≤[m(m +1)]^n.But the product of all pairs is exactly (m +n)! / (m -n)!.Hence, (m +n)! / (m -n)! = product_{k=1 to n} (m +n -k +1)(m -n +k) ≤ [m(m +1)]^n = (m² +m)^n.Therefore, the right inequality holds.Wait, that seems straightforward. Each pair is ≤m(m +1), so multiplying n pairs gives [m(m +1)]^n. Therefore, the right inequality is proven.So, summarizing:For the left inequality, used induction, pairing terms, and showed that each induction step requires the product to grow sufficiently.For the right inequality, noticed that each pair of terms in the product is bounded above by m(m +1), so multiplying n such pairs gives the upper bound.Therefore, both inequalities hold, hence the result.Final AnswerThe inequalities are proven to hold, so the solution is boxed{2^n n! leq frac{(m+n)!}{(m-n)!} leq (m^2 + m)^n}.

✅:To prove the inequality (2^n n! leq frac{(m+n)!}{(m-n)!} leq (m^2 + m)^n) for positive integers (m) and (n) with (n leq m), we consider each part of the inequality separately. Left Inequality: (2^n n! leq frac{(m+n)!}{(m-n)!})1. Understanding the Product: - The expression (frac{(m+n)!}{(m-n)!}) represents the product of (2n) consecutive integers from (m-n+1) to (m+n). - This product can be paired into (n) pairs: ((m+n)(m-n+1), (m+n-1)(m-n+2), ldots, (m+1)m).2. Inductive Proof: - Base Case: For (n=1), (2 leq (m+1)m), which holds since (m geq 1). - Inductive Step: Assume it holds for (n=k). For (n=k+1), we need to show ((m+k+1)(m-k) geq 2(k+1)). This is true since ((m+k+1)(m-k) geq 2(k+1)) when (m geq k+1). Right Inequality: (frac{(m+n)!}{(m-n)!} leq (m^2 + m)^n)1. Bounding Each Pair: - Each pair ((m+n-k+1)(m-n+k)) in the product (frac{(m+n)!}{(m-n)!}) is bounded above by (m(m+1)). - For each pair, ((m+n-k+1)(m-n+k) leq m(m+1)) because ((m+n-k+1)(m-n+k) = m^2 + m - n(n-1) leq m(m+1)).2. Product of Pairs: - Since there are (n) pairs, the product of all pairs is (left(m(m+1)right)^n = (m^2 + m)^n). ConclusionBoth inequalities are proven by induction and bounding each pair in the product. Therefore, the final result is:[boxed{2^n n! leq frac{(m+n)!}{(m-n)!} leq (m^2 + m)^n}]

🔑:1. Prove the left inequality: [ 2^n n! leq frac{(m+n)!}{(m-n)!} ] We start by expanding the factorials: [ frac{(m+n)!}{(m-n)!} = (m+n)(m+n-1)cdots(m-n+1) ] Since ( n leq m ), each term in the product ( (m+n)(m+n-1)cdots(m-n+1) ) is greater than or equal to ( n+1 ). Therefore, we can write: [ (m+n)(m+n-1)cdots(m-n+1) geq n(n-1)cdots1 = n! ] To show that this product is at least ( 2^n n! ), we note that: [ (m+n)(m+n-1)cdots(m-n+1) geq 2^n n! ] This can be seen by induction or by noting that each term ( m+k ) (where ( k ) ranges from ( -n ) to ( n )) is at least 2. Thus: [ frac{(m+n)!}{(m-n)!} geq 2^n n! ]2. Prove the right inequality: [ frac{(m+n)!}{(m-n)!} leq (m^2 + m)^n ] We start by expanding the factorials: [ frac{(m+n)!}{(m-n)!} = (m+n)(m+n-1)cdots(m-n+1) ] We need to show that: [ (m+n)(m+n-1)cdots(m-n+1) leq (m^2 + m)^n ] Consider the product of terms: [ (m+n)(m+n-1)cdots(m-n+1) ] Each term ( m+k ) (where ( k ) ranges from ( -n ) to ( n )) is at most ( m+n ). Therefore, we can bound each term by ( m^2 + m ): [ (m+k) leq m^2 + m ] Since there are ( n ) terms in the product, we have: [ (m+n)(m+n-1)cdots(m-n+1) leq (m^2 + m)^n ]The final answer is ( boxed{ 2^n n! leq frac{(m+n)!}{(m-n)!} leq (m^2 + m)^n } )

🧠:Okay, so I have this problem where for each nonnegative integer n, we define A_n = 2^{3n} + 3^{6n+2} + 5^{6n+2}. I need to find the greatest common divisor (GCD) of all the numbers A_0, A_1, ..., A_{1999}. Hmm, that sounds like I need to find a number that divides every A_n for n from 0 to 1999, and it's the largest such number. First, maybe I should start by computing the first few terms to see if I can spot a pattern or figure out possible common divisors. Let's start with n=0, n=1, n=2, and see what those A_n values are.For n=0:A_0 = 2^{0} + 3^{2} + 5^{2} = 1 + 9 + 25 = 35.For n=1:A_1 = 2^{3} + 3^{8} + 5^{8} = 8 + 6561 + 390625. Let's compute that:8 + 6561 = 6569; 6569 + 390625 = 397194. Wait, 390625 + 6561 is 397186, plus 8 is 397194. Hmm, that's 397194.For n=2:A_2 = 2^{6} + 3^{14} + 5^{14}. Let's calculate each term:2^6 = 643^14 is 47829695^14 is 6103515625Adding them up: 64 + 4,782,969 = 4,783,033; then 4,783,033 + 6,103,515,625 = 6,108,298,658.So, A_0 = 35, A_1 = 397,194, A_2 = 6,108,298,658. Now, I need to find GCD of these numbers. The GCD of all A_n would have to divide each of them, so let's check the GCD of A_0 and A_1 first.Compute GCD(35, 397194). Let's see:First, divide 397194 by 35. 35*11,000 = 385,000. 397,194 - 385,000 = 12,194. 35*348 = 12,180. So 12,194 - 12,180 = 14. So 397194 = 35*11348 + 14. Therefore, GCD(35,14) = 7. Then, GCD(7, 14) is 7. So the GCD of 35 and 397194 is 7.Now check if 7 divides A_2. Let's compute A_2 modulo 7.First, 2^6 mod7: 2^3=8≡1 mod7, so 2^6=(2^3)^2≡1^2=1 mod7.3^14 mod7: Since 3^6≡1 mod7 by Fermat's little theorem, so 3^14=3^(6*2 + 2)= (3^6)^2 *3^2 ≡1^2 *9 ≡2 mod7.5^14 mod7: 5≡5 mod7. 5^6≡1 mod7, so 5^14=5^(6*2 +2)= (5^6)^2 *5^2 ≡1^2 *25≡4 mod7.So A_2 mod7=1 + 2 +4=7≡0 mod7. Therefore, 7 divides A_2. So 7 is a common divisor so far.Wait, so GCD(A_0, A_1, A_2) is 7. Let's check if there's a higher common divisor. Suppose we check 14. Let's see if 14 divides A_0=35. 35 divided by 14 is 2.5, which is not an integer, so 14 doesn't divide 35. So GCD can't be 14. Similarly, 35's divisors are 1,5,7,35. Since 7 is the only common divisor so far. Let's check A_3 modulo 7 to see if 7 continues to divide all A_n.But perhaps we can proceed more formally. Let's try to compute A_n modulo 7 for arbitrary n.Given that A_n = 2^{3n} + 3^{6n+2} +5^{6n+2}Compute each term modulo 7.First term: 2^{3n} mod7. Let's note that 2^3=8≡1 mod7. Therefore, 2^{3n}=(2^3)^n ≡1^n=1 mod7. So 2^{3n}≡1 mod7.Second term: 3^{6n+2} mod7. Since 3^6≡1 mod7 by Fermat's little theorem (since 7 is prime and 3≠0 mod7). Therefore, 3^{6n+2}=3^{6n}*3^2≡(1)^n*9≡2 mod7.Third term:5^{6n+2} mod7. Similarly, 5^6≡1 mod7, so 5^{6n+2}=5^{6n}*5^2≡1^n*25≡4 mod7.Therefore, adding them up: 1 + 2 +4=7≡0 mod7. Therefore, for any n, A_n ≡0 mod7. Thus 7 divides every A_n.So 7 is a common divisor. Now, we need to check if there's a larger common divisor. Since A_0=35, the possible divisors are 1,5,7,35. But 35 divides A_0=35, but does 35 divide A_1=397194? Let's check.397194 divided by 35: 35*10,000=350,000. 397,194-350,000=47,194. 35*1,000=35,000. 47,194-35,000=12,194. Then, as before, 35*348=12,180. 12,194-12,180=14. So remainder 14. So 35 does not divide A_1. Thus 35 is out. Then, the GCD could be 7 or 5. Let's check if 5 divides all A_n.Check A_0=35: 35 is divisible by 5. A_1=397,194: last digit is 4, so not divisible by 5. Therefore, 5 cannot be a common divisor. Therefore, the GCD is 7. Wait, but let me confirm that 7 is indeed the greatest common divisor.Alternatively, maybe 7 is the GCD. But let's check if any higher prime divisor divides all A_n. Let's check A_0=35. Its prime factors are 5 and 7. Since 5 doesn't divide A_1, only possible common prime factor is 7. Therefore, the GCD is 7. Hence, the answer should be 7. But let me check another term, like A_3, just to be thorough.Compute A_3 modulo 7. As above, for any n, each term is 1 +2 +4=7≡0 mod7. So yes, 7 divides all A_n, and since no higher number divides all A_n, the GCD is 7. Therefore, the answer is 7. But let me check with n=1. A_1=397,194. Divided by 7: 7*56,000=392,000. 397,194-392,000=5,194. 7*700=4,900. 5,194-4,900=294. 7*42=294. So total is 56,000+700+42=56,742. 56,742*7=397,194. Correct. So 7 divides A_1. Then A_2 as we saw earlier is 6,108,298,658. Divided by 7: 7*872,614,094=6,108,298,658. So yes. Therefore, 7 is indeed a common divisor. Since there's no higher common divisor, the GCD is 7.But wait, let me check A_0=35, which is 5*7. If there was another prime factor common to all A_n, it would have to divide 35, so 5 or 7. But 5 doesn't divide A_1, as we saw. So only 7. Therefore, the GCD is 7.Alternatively, maybe 7 is the only common divisor. Therefore, the answer is 7. But let me see if maybe there's a mistake in my reasoning. For example, perhaps all the terms are congruent modulo some higher number. Suppose we check modulo 14. Let's check A_0=35≡7 mod14. A_1=397,194 mod14: 397,194 divided by 14: 14*28,371=397,194. So 397,194≡0 mod14? Wait, 14*28,371=397,194. So 397,194 is divisible by 14. Then A_0=35≡7 mod14, which is not 0. So 14 does not divide A_0. Therefore, 14 is not a common divisor. Similarly, any multiple of 7 and another prime would not divide A_0=35. Since 35=5*7, and 5 isn't a divisor for all terms, so 7 is the maximum.Alternatively, maybe 7^2=49 divides some A_n. Let's check A_0=35. 35 mod49=35, so no. Therefore, 49 cannot be a divisor. Therefore, 7 is the maximum.Thus, the greatest common divisor is 7. So the answer is boxed{7}.Final Answerboxed{7}

✅:To find the greatest common divisor (GCD) of the numbers (A_0, A_1, ldots, A_{1999}) where (A_n = 2^{3n} + 3^{6n+2} + 5^{6n+2}), we start by computing the first few terms and analyzing their divisibility.First, we calculate (A_0), (A_1), and (A_2):- For (n=0): [ A_0 = 2^0 + 3^2 + 5^2 = 1 + 9 + 25 = 35 ]- For (n=1): [ A_1 = 2^3 + 3^8 + 5^8 = 8 + 6561 + 390625 = 397194 ]- For (n=2): [ A_2 = 2^6 + 3^{14} + 5^{14} = 64 + 4782969 + 6103515625 = 6108298658 ]Next, we compute the GCD of (A_0) and (A_1):- (text{GCD}(35, 397194)): Using the Euclidean algorithm, we find (text{GCD}(35, 14) = 7).We then check if 7 divides (A_2) by computing each term modulo 7:- (2^6 equiv 1 mod 7)- (3^{14} equiv (3^6)^2 cdot 3^2 equiv 1^2 cdot 2 equiv 2 mod 7)- (5^{14} equiv (5^6)^2 cdot 5^2 equiv 1^2 cdot 4 equiv 4 mod 7)- Summing these: (1 + 2 + 4 = 7 equiv 0 mod 7)Thus, 7 divides (A_2). To confirm 7 divides all (A_n), we analyze (A_n) modulo 7 for general (n):- (2^{3n} equiv (2^3)^n equiv 1^n equiv 1 mod 7)- (3^{6n+2} equiv (3^6)^n cdot 3^2 equiv 1^n cdot 2 equiv 2 mod 7)- (5^{6n+2} equiv (5^6)^n cdot 5^2 equiv 1^n cdot 4 equiv 4 mod 7)- Summing these: (1 + 2 + 4 = 7 equiv 0 mod 7)Since 7 divides all (A_n) and no higher number (like 14 or 35) divides all terms, the GCD is 7. Therefore, the greatest common divisor of (A_0, A_1, ldots, A_{1999}) is (boxed{7}).

🔑:1. First, we define the sequence ( A_n = 2^{3n} + 3^{6n+2} + 5^{6n+2} ) for nonnegative integers ( n ).2. We need to find the greatest common divisor (GCD) of the numbers ( A_0, A_1, ldots, A_{1999} ).3. Calculate ( A_0 ): [ A_0 = 2^{3 cdot 0} + 3^{6 cdot 0 + 2} + 5^{6 cdot 0 + 2} = 2^0 + 3^2 + 5^2 = 1 + 9 + 25 = 35 ] So, ( A_0 = 35 ).4. Next, we evaluate ( A_1 ) modulo 5 and modulo 7.5. Evaluate ( A_1 mod 5 ): [ A_1 = 2^{3 cdot 1} + 3^{6 cdot 1 + 2} + 5^{6 cdot 1 + 2} = 2^3 + 3^8 + 5^8 ] [ 2^3 equiv 3 pmod{5} ] By Fermat's Little Theorem, ( 3^4 equiv 1 pmod{5} ), so: [ 3^8 equiv (3^4)^2 equiv 1^2 equiv 1 pmod{5} ] [ 5^8 equiv 0 pmod{5} ] Therefore: [ A_1 equiv 3 + 1 + 0 equiv 4 pmod{5} ]6. Evaluate ( A_1 mod 7 ): [ A_1 = 2^{3 cdot 1} + 3^{6 cdot 1 + 2} + 5^{6 cdot 1 + 2} = 2^3 + 3^8 + 5^8 ] [ 2^3 equiv 1 pmod{7} ] By Fermat's Little Theorem, ( 3^6 equiv 1 pmod{7} ), so: [ 3^8 equiv 3^6 cdot 3^2 equiv 1 cdot 9 equiv 2 pmod{7} ] [ 5^8 equiv (-2)^8 equiv 2^8 equiv 256 equiv 4 pmod{7} ] Therefore: [ A_1 equiv 1 + 2 + 4 equiv 7 equiv 0 pmod{7} ]7. We now generalize the result for all ( n ): [ A_n = 2^{3n} + 3^{6n+2} + 5^{6n+2} ] We need to show ( A_n equiv 0 pmod{7} ) for all ( n ).8. Consider ( 2^{3n} mod 7 ): [ 2^{3n} equiv 1 pmod{7} quad text{(since } 2^3 equiv 1 pmod{7}) ]9. Consider ( 3^{6n+2} mod 7 ): [ 3^{6n+2} equiv 3^2 equiv 9 equiv 2 pmod{7} quad text{(since } 3^6 equiv 1 pmod{7}) ]10. Consider ( 5^{6n+2} mod 7 ): [ 5^{6n+2} equiv (-2)^{6n+2} equiv 2^{6n+2} equiv 4 pmod{7} quad text{(since } 2^3 equiv 1 pmod{7}) ]11. Summing these results: [ A_n equiv 1 + 2 + 4 equiv 7 equiv 0 pmod{7} ]12. Since ( A_n equiv 0 pmod{7} ) for all ( n ), the greatest common divisor of ( A_0, A_1, ldots, A_{1999} ) is 7.The final answer is (boxed{7}).