🧠:Okay, so I need to find the value of the function f(x) = cos(2x) at x=0, as well as its first four derivatives evaluated at x=0. Let me start by recalling some basic calculus rules, especially those related to trigonometric functions and derivatives. First off, the original function is cosine of 2x. I know that the derivative of cosine is negative sine, and then there's the chain rule to consider because of the 2x inside the cosine. The chain rule says that the derivative of cos(u) with respect to x is -sin(u) times the derivative of u with respect to x. So in this case, u = 2x, so the derivative of u is 2. That means each time I take a derivative, I should expect a factor of 2 coming out from the chain rule, and the trigonometric function will alternate between cosine and sine, with alternating signs as well. Let me try to methodically compute each derivative step by step.Starting with f(x) = cos(2x). Evaluating this at x=0 should be straightforward. cos(0) is 1, because cos(0) = 1. But wait, the argument here is 2x, so when x=0, it's cos(0) which is indeed 1. So f(0) = 1.Now, moving on to the first derivative, f'(x). The derivative of cos(u) is -sin(u)*u', so here u = 2x, so u' = 2. Therefore, f'(x) = -sin(2x)*2 = -2 sin(2x). Now, evaluate this at x=0. sin(0) is 0, so f'(0) = -2*0 = 0. That seems right.Next, the second derivative, f''(x). Starting from f'(x) = -2 sin(2x). The derivative of sin(u) is cos(u)*u', so here the derivative of sin(2x) is cos(2x)*2. So applying that, f''(x) = derivative of -2 sin(2x) which is -2 * [cos(2x)*2] = -4 cos(2x). Then evaluating at x=0, cos(0) is 1, so f''(0) = -4 * 1 = -4. Wait a second, but hold on. Let me check that again. The derivative of sin(2x) is 2 cos(2x), so multiplying by the -2 from f'(x), it's -2 * 2 cos(2x) = -4 cos(2x). Yes, that's correct. So f''(0) = -4 * 1 = -4. Okay.Third derivative, f'''(x). Starting from f''(x) = -4 cos(2x). The derivative of cos(2x) is -sin(2x)*2, so f'''(x) = derivative of -4 cos(2x) = -4 * [-sin(2x)*2] = 8 sin(2x). Then evaluating at x=0, sin(0) is 0, so f'''(0) = 8 * 0 = 0. That seems straightforward.Fourth derivative, f''''(x) or f⁗(x). Starting from f'''(x) = 8 sin(2x). The derivative of sin(2x) is 2 cos(2x), so f''''(x) = 8 * 2 cos(2x) = 16 cos(2x). Then evaluating at x=0, cos(0) = 1, so f⁗(0) = 16 * 1 = 16. Wait a moment. Let me verify each step again to make sure I didn't mix up any signs or factors.Starting with f(x) = cos(2x):1. f(0) = cos(0) = 1. Correct.2. First derivative: f'(x) = -2 sin(2x). At x=0: 0. Correct.3. Second derivative: derivative of -2 sin(2x) is -4 cos(2x). At x=0: -4. Correct.4. Third derivative: derivative of -4 cos(2x) is 8 sin(2x). At x=0: 0. Correct.5. Fourth derivative: derivative of 8 sin(2x) is 16 cos(2x). At x=0: 16. Correct.Hmm, so the pattern here seems to be that each derivative introduces a factor of 2 and alternates between sine and cosine, with the sign alternating every two derivatives. Let me see:First derivative: -2 sin(2x) (factor of 2, negative sign)Second derivative: -4 cos(2x) (another factor of 2, stays negative because derivative of sine is cosine, but original coefficient was negative, so derivative is -2*2 cos = -4 cos)Third derivative: 8 sin(2x) (another factor of 2, sign becomes positive because derivative of -4 cos is 8 sin)Fourth derivative: 16 cos(2x) (another factor of 2, sign stays positive because derivative of sin is cos, multiplied by 8*2=16)Yes, that makes sense. Each derivative multiplies by 2 and alternates the trigonometric function. The sign alternates every other derivative because:- The first derivative is -2 sin, which is negative.- The second derivative is -4 cos, still negative.- The third derivative becomes positive because the derivative of cos is -sin, and we have -4*(-sin*2) = +8 sin.- The fourth derivative is derivative of sin, which is cos*2, so 8*2 cos = 16 cos, positive.Therefore, evaluated at x=0, the even derivatives (0th, 2nd, 4th) will involve cos(0) = 1, while the odd derivatives (1st, 3rd) involve sin(0) = 0. Hence, the odd derivatives at 0 are zero, and the even derivatives alternate in sign? Wait, but looking at the results:f(0) = 1 (0th derivative)f''(0) = -4f⁗(0) = 16So the signs are alternating starting from positive at 0th derivative, then negative at 2nd, positive at 4th. So the pattern is: for even derivatives, the sign is (-1)^{n/2} where n is the derivative order. Let me check:0th derivative: (-1)^0 = 1, matches 1.2nd derivative: (-1)^1 = -1, so -4 = (-1)^1 * 4, but here we have -4 which is (-1)^1 * 4. Wait, but 4 is (2)^2.Similarly, 4th derivative: (-1)^2 * (2)^4 = 1 * 16. But 16 is 2^4. Wait, so maybe the general formula for the nth derivative at 0 is:If n is even, then f^{(n)}(0) = (-1)^{n/2} * (2)^n.If n is odd, then f^{(n)}(0) = 0.Let me verify with our results:n=0 (even): (-1)^0 * 2^0 = 1*1=1. Correct.n=1 (odd): 0. Correct.n=2 (even): (-1)^1 * 2^2 = -1*4 = -4. Correct.n=3 (odd): 0. Correct.n=4 (even): (-1)^2 * 2^4 = 1*16 =16. Correct.Yes, that seems to hold. Therefore, the even derivatives at 0 follow (-1)^{k} * (2)^{2k} where n=2k. Wait, let's see:If n is even, say n=2k, then the derivative is (-1)^k * (2)^{2k} ?Wait, n=2k, so k = n/2.Therefore, f^{(n)}(0) = (-1)^{n/2} * (2)^n.Wait, for n=2, that would be (-1)^1 * 4 = -4. Correct.n=4: (-1)^2 * 16 = 16. Correct.So indeed, the formula is for even n: f^{(n)}(0) = (-1)^{n/2} * (2)^n. And for odd n, it's 0.Therefore, the values we found make sense in the general pattern.Alternatively, another way to look at this is through the Taylor series expansion of cos(2x) around x=0. The Taylor series for cos(u) is Σ_{k=0}^∞ (-1)^k u^{2k}/(2k)!). So substituting u = 2x, we get:cos(2x) = Σ_{k=0}^∞ (-1)^k (2x)^{2k}/(2k)! ) = Σ_{k=0}^∞ (-1)^k 2^{2k} x^{2k}/(2k)! )Therefore, the coefficients of x^n in the Taylor series are:For even n=2k: (-1)^k 2^{2k}/(2k)! )For odd n: 0.But the Taylor series is also Σ_{n=0}^∞ f^{(n)}(0)/n! x^n.Therefore, comparing coefficients, for even n=2k:f^{(2k)}(0)/(2k)! = (-1)^k 2^{2k}/(2k)! )Therefore, f^{(2k)}(0) = (-1)^k 2^{2k}Which is the same as before. For example, when k=0 (n=0): f(0) = (-1)^0 2^0 =1.k=1 (n=2): (-1)^1 2^2 = -4.k=2 (n=4): (-1)^2 2^4 = 16.So yes, this confirms the results we obtained by directly computing the derivatives.Therefore, the values are:f(0) =1f'(0)=0f''(0)=-4f'''(0)=0f''''(0)=16Let me just recap once more to ensure I didn't miscalculate any derivatives.First derivative:d/dx [cos(2x)] = -2 sin(2x). At 0: 0. Correct.Second derivative:d/dx [-2 sin(2x)] = -2 * 2 cos(2x) = -4 cos(2x). At 0: -4. Correct.Third derivative:d/dx [-4 cos(2x)] = -4 * (-2 sin(2x)) = 8 sin(2x). At 0: 0. Correct.Fourth derivative:d/dx [8 sin(2x)] = 8 * 2 cos(2x) = 16 cos(2x). At 0:16. Correct.All steps check out. Each time, the coefficient multiplies by 2, and the trigonometric function toggles between sin and cos, with the sign changing every other derivative. The zeros at the odd derivatives make sense because cosine is an even function, so all its odd derivatives, which are odd functions, must be zero at x=0.Alternatively, since cos(2x) is an even function, all its odd derivatives are odd functions, and odd functions evaluated at 0 are 0. That's another way to see why f'(0), f'''(0), etc., are zero.Therefore, the computed values are consistent with the properties of even and odd functions as well.Another way to verify is to use the Maclaurin series expansion of cos(2x):cos(2x) = 1 - (2x)^2/2! + (2x)^4/4! - (2x)^6/6! + ...Which simplifies to:1 - 4x^2/2 + 16x^4/24 - 64x^6/720 + ...= 1 - 2x^2 + (2/3)x^4 - (4/45)x^6 + ...From this series, the coefficients of x^0, x^1, x^2, x^3, x^4, etc., are:1, 0, -2, 0, 2/3, 0, -4/45, ...But wait, the derivatives at 0 are related to the coefficients by f^{(n)}(0) = n! * a_n, where a_n is the coefficient of x^n.Therefore:f(0) = 0! * a_0 = 1 *1 =1.f'(0) =1! * a_1 =1*0=0.f''(0)=2! *a_2=2*(-2)= -4.f'''(0)=3! *a_3=6*0=0.f''''(0)=4! *a_4=24*(2/3)=24*(2/3)=16.Yes, exactly as we found earlier. So this cross-verification using the Taylor series also confirms that the derivatives at 0 are correct.Therefore, after thoroughly checking through multiple methods—direct computation, considering the pattern of derivatives, using properties of even functions, and verifying via the Taylor series expansion—I can confidently say that the values are accurate.Final AnswerThe values are f(0) = boxed{1}, f'(0) = boxed{0}, f''(0) = boxed{-4}, f'''(0) = boxed{0}, and f^{mathrm{IV}}(0) = boxed{16}.

✅:To find ( f(0), f'(0), f''(0), f'''(0), f^{mathrm{IV}}(0) ) for ( f(x) = cos(2x) ):1. Evaluating ( f(0) ): [ f(0) = cos(2 cdot 0) = cos(0) = 1 ]2. First derivative ( f'(x) ): [ f'(x) = frac{d}{dx} cos(2x) = -2 sin(2x) ] Evaluating at ( x = 0 ): [ f'(0) = -2 sin(0) = 0 ]3. Second derivative ( f''(x) ): [ f''(x) = frac{d}{dx} (-2 sin(2x)) = -4 cos(2x) ] Evaluating at ( x = 0 ): [ f''(0) = -4 cos(0) = -4 ]4. Third derivative ( f'''(x) ): [ f'''(x) = frac{d}{dx} (-4 cos(2x)) = 8 sin(2x) ] Evaluating at ( x = 0 ): [ f'''(0) = 8 sin(0) = 0 ]5. Fourth derivative ( f^{mathrm{IV}}(x) ): [ f^{mathrm{IV}}(x) = frac{d}{dx} (8 sin(2x)) = 16 cos(2x) ] Evaluating at ( x = 0 ): [ f^{mathrm{IV}}(0) = 16 cos(0) = 16 ] Final AnswerThe values are ( f(0) = boxed{1} ), ( f'(0) = boxed{0} ), ( f''(0) = boxed{-4} ), ( f'''(0) = boxed{0} ), and ( f^{mathrm{IV}}(0) = boxed{16} ).

🔑:To find ( f(0), f'(0), f''(0), f'''(0), f^{mathrm{IV}}(0) ) given that ( f(x) = cos(2x) ), we will compute the derivatives of ( f(x) ) up to the fourth order and then evaluate these derivatives at ( x = 0 ).1. Compute the first derivative: [ f'(x) = frac{d}{dx} cos(2x) ] Using the chain rule: [ f'(x) = -sin(2x) cdot frac{d}{dx} (2x) = -2 sin(2x) ] Therefore: [ f'(x) = -2 sin(2x) ]2. Compute the second derivative: [ f''(x) = frac{d}{dx} f'(x) = frac{d}{dx} (-2 sin(2x)) ] Again, applying the chain rule: [ f''(x) = -2 cos(2x) cdot frac{d}{dx} (2x) = -4 cos(2x) ] Therefore: [ f''(x) = -4 cos(2x) ]3. Compute the third derivative: [ f'''(x) = frac{d}{dx} f''(x) = frac{d}{dx} (-4 cos(2x)) ] Applying the chain rule: [ f'''(x) = -4 (-sin(2x)) cdot frac{d}{dx} (2x) = 8 sin(2x) ] Therefore: [ f'''(x) = 8 sin(2x) ]4. Compute the fourth derivative: [ f^{mathrm{IV}}(x) = frac{d}{dx} f'''(x) = frac{d}{dx} (8 sin(2x)) ] Applying the chain rule: [ f^{mathrm{IV}}(x) = 8 cos(2x) cdot frac{d}{dx} (2x) = 16 cos(2x) ] Therefore: [ f^{mathrm{IV}}(x) = 16 cos(2x) ]5. Evaluate the derivatives at ( x = 0 ): - ( f(0) = cos(2 cdot 0) = cos(0) = 1 ) - ( f'(0) = -2 sin(2 cdot 0) = -2 sin(0) = 0 ) - ( f''(0) = -4 cos(2 cdot 0) = -4 cos(0) = -4 ) - ( f'''(0) = 8 sin(2 cdot 0) = 8 sin(0) = 0 ) - ( f^{mathrm{IV}}(0) = 16 cos(2 cdot 0) = 16 cos(0) = 16 ) Conclusion:[boxed{f(0)=1, f^{prime}(0)=0, f^{prime prime}(0)=-4, f^{prime prime prime}(0)=0, f^{mathrm{IV}}(0)=16}]

🧠:Okay, so I have this problem here: Given that (y + z)/x = (z + x)/y = (x + y)/z = k, find the value of k. The options are A) 2 or -1, B) 2, C) -1, D) 2 or 1. Hmm, let me try to figure this out step by step.First, the problem states that three different ratios are all equal to the same constant k. That means:(y + z)/x = k (z + x)/y = k (x + y)/z = kSo each of these fractions equals k. Maybe I can start by writing each equation separately and see if I can find relationships between x, y, and z.Starting with the first equation: (y + z)/x = k. Multiplying both sides by x gives y + z = kx.Similarly, the second equation: (z + x)/y = k. Multiplying by y gives z + x = ky.Third equation: (x + y)/z = k. Multiplying by z gives x + y = kz.Now, I have three equations:1. y + z = kx 2. z + x = ky 3. x + y = kzHmm, so three equations with variables x, y, z and the constant k. Since we need to find k, perhaps we can express x, y, z in terms of each other and find a relationship that leads to k.Let me try to express each variable in terms of the others. Let's take equation 1: y + z = kx. Maybe express x here:x = (y + z)/kSimilarly, from equation 2: z + x = ky. Let's express y here:y = (z + x)/kAnd from equation 3: x + y = kz. Express z here:z = (x + y)/kWait, this seems cyclic. Each variable is expressed in terms of the sum of the other two divided by k. Maybe substituting these into each other will help.Alternatively, perhaps add all three equations together. Let's try that.Adding equations 1, 2, and 3:(y + z) + (z + x) + (x + y) = kx + ky + kzSimplify the left side:y + z + z + x + x + y = 2x + 2y + 2z = 2(x + y + z)Right side: k(x + y + z)So, 2(x + y + z) = k(x + y + z)Hmm, that's interesting. Let's denote S = x + y + z. Then the equation becomes:2S = kSIf S ≠ 0, we can divide both sides by S:2 = kBut if S = 0, then the equation 2S = kS becomes 0 = 0, which is always true, but k could be anything? Wait, no. Wait, but in that case, if S = 0, we have x + y + z = 0, so we need to check what k would be in that scenario.So, there are two cases: either S ≠ 0, leading to k = 2, or S = 0, which we need to explore further.Let me handle each case separately.Case 1: S ≠ 0. Then k = 2. Let me check if this is possible.If k = 2, then from the original equations:y + z = 2x z + x = 2y x + y = 2zLet me see if there's a solution here. For example, suppose all variables are equal. Let’s assume x = y = z. Then, substituting into the first equation: x + x = 2x ⇒ 2x = 2x, which is true. Similarly, all equations hold. So x = y = z is a solution when k = 2. So that works. So k = 2 is a valid solution.Case 2: S = 0. Then x + y + z = 0. Let's see what k is in this case.From the original equations, since x + y + z = 0, maybe we can express one variable in terms of the others. Let's say z = -x - y.Substitute z = -x - y into the first equation: (y + z)/x = k.So substituting z:(y + (-x - y))/x = (-x)/x = -1 = k. So k = -1.Similarly, check the second equation: (z + x)/y = k. Substitute z:(-x - y + x)/y = (-y)/y = -1 = k.Third equation: (x + y)/z = k. Since z = -x - y, then (x + y)/z = (x + y)/(-x - y) = -1 = k.So in this case, all three equations give k = -1. Therefore, when x + y + z = 0, k must be -1.Therefore, k can be either 2 or -1. So the possible values of k are 2 and -1. Therefore, the correct answer should be A) 2 or -1.But wait, let me verify if there's any restriction. Suppose all x, y, z are zero. Then the original expressions would be undefined because we can't divide by zero. So we must assume that x, y, z are not all zero. However, if x + y + z = 0, it's possible that some variables are non-zero but their sum is zero, so that's acceptable as long as none of x, y, z are individually zero. Wait, but if x is zero, then (y + z)/x would be undefined, so we must have x, y, z all non-zero. So in the problem statement, implicitly, x, y, z ≠ 0.Therefore, in Case 2, when x + y + z = 0, none of x, y, z can be zero. So this is a valid scenario.Therefore, there are two possibilities: either k = 2 (when x + y + z ≠ 0) or k = -1 (when x + y + z = 0). Both are valid, so the answer is A) 2 or -1.But let me check if the problem allows both. The answer choices include A) 2 or -1, which would be the correct answer based on this reasoning. But let me check again.Alternatively, maybe I made a mistake. Let's test with actual numbers.First, test k = 2. Let x = y = z = 1. Then:(y + z)/x = (1 + 1)/1 = 2 (z + x)/y = (1 + 1)/1 = 2 (x + y)/z = (1 + 1)/1 = 2 So k = 2 works.Now test k = -1. Let’s have x + y + z = 0. For example, let x = 1, y = 1, then z = -2.Check (y + z)/x = (1 + (-2))/1 = (-1)/1 = -1 (z + x)/y = (-2 + 1)/1 = (-1)/1 = -1 (x + y)/z = (1 + 1)/(-2) = 2/(-2) = -1 So yes, k = -1 works here.Therefore, both k = 2 and k = -1 are possible. Hence, answer A is correct.But wait, the options are given as (A) 2 or -1, (B) 2, (C) -1, (D) 2 or 1. So since both 2 and -1 are possible, the answer should be A. So the correct answer is A.But let me check again if there's a case where k cannot be -1. Suppose someone argues that x, y, z must be positive numbers. But the problem didn't specify that. If negative numbers are allowed, then k = -1 is possible. The problem doesn't state any restrictions on x, y, z, other than being non-zero (since denominators can't be zero). So if x, y, z can be negative, then k can be -1. So both values are valid. Therefore, the answer is A.Alternatively, maybe there's a trick here. Let's consider if there's another approach. Suppose we take the equations:y + z = kx z + x = ky x + y = kzLet me subtract the second equation from the first. First equation: y + z = kx Second equation: z + x = ky Subtracting second from first: (y + z) - (z + x) = kx - ky Simplify: y - x = k(x - y) Factor: (y - x) = -k(y - x)So (y - x)(1 + k) = 0.Therefore, either y = x or 1 + k = 0, i.e., k = -1.Similarly, subtracting the third equation from the second:Second equation: z + x = ky Third equation: x + y = kz Subtract: (z + x) - (x + y) = ky - kz Simplify: z - y = k(y - z) Factor: (z - y) = -k(z - y) So (z - y)(1 + k) = 0.Thus, either z = y or k = -1.Similarly, subtracting first from third:Third equation: x + y = kz First equation: y + z = kx Subtract: (x + y) - (y + z) = kz - kx Simplify: x - z = k(z - x) Factor: (x - z) = -k(x - z) Thus, (x - z)(1 + k) = 0.So again, either x = z or k = -1.Therefore, from these three subtractions, we get that either all variables are equal (x = y = z) or k = -1.If x = y = z, then substituting back into the original equations:(y + z)/x = (x + x)/x = 2x/x = 2 = k. So k = 2.Alternatively, if k = -1, then as we saw before, the equations hold when x + y + z = 0.Thus, the possible values of k are 2 or -1, so answer A.Therefore, the correct answer is A) 2 or -1.Final Answerboxed{A}

✅:Given the equations (frac{y+z}{x} = frac{z+x}{y} = frac{x+y}{z} = k), we need to find the value of (k).First, we express each equation separately:1. (y + z = kx)2. (z + x = ky)3. (x + y = kz)Adding all three equations, we get:[(y + z) + (z + x) + (x + y) = kx + ky + kz]Simplifying the left side:[2(x + y + z) = k(x + y + z)]Let (S = x + y + z). Then:[2S = kS]If (S neq 0), we can divide both sides by (S) to get (k = 2).If (S = 0), then (x + y + z = 0). Substituting (z = -x - y) into the first equation:[frac{y + (-x - y)}{x} = frac{-x}{x} = -1 = k]Similarly, substituting into the other equations also gives (k = -1).To verify, we check both cases:- For (k = 2), if (x = y = z), all equations hold.- For (k = -1), if (x + y + z = 0), all equations hold.Subtracting pairs of equations, we find that either (x = y = z) (leading to (k = 2)) or (k = -1).Thus, the possible values of (k) are 2 or -1, leading to the answer (boxed{A}).

🔑:Given the system of equations:frac{y+z}{x} = frac{z+x}{y} = frac{x+y}{z} = k,1. We start by recognizing that the given ratios are all equal to the constant k. This implies: [ frac{y+z}{x} = k, quad frac{z+x}{y} = k, quad frac{x+y}{z} = k ]2. We can solve one of the equations for k. Let’s take frac{y+z}{x} = k and cross-multiply: [ y + z = kx ]3. Similarly, from frac{z+x}{y} = k, we get: [ z + x = ky ]4. Finally, from frac{x+y}{z} = k, we get: [ x + y = kz ]5. Adding all these equations together: [ (y + z) + (z + x) + (x + y) = kx + ky + kz ]6. Combine like terms on the left-hand side and factor out k on the right-hand side: [ 2(x + y + z) = k(x + y + z) ]7. If x + y + z neq 0, we can divide both sides by (x + y + z): [ 2 = k ]8. If x + y + z = 0, substitute y + z = -x back into the first equation to find: [ frac{y+z}{x} = frac{-x}{x} = -1 implies k = -1 ] Conclusion:Thus, the possible values for k are 2 and -1.boxed{A}

🧠:Okay, so I need to show that if I slice an oblate spheroid (which is formed by rotating an ellipse around its minor axis) with a plane that's parallel to the axis of rotation, the resulting cross-section is an ellipse. Hmm, let me start by recalling what an oblate spheroid is and how it's formed.First, an ellipse in the standard form is (x²/a²) + (y²/b²) = 1. If we rotate this ellipse around its minor axis, which is the y-axis if b < a, then we get an oblate spheroid. The equation for the spheroid would then be (x² + z²)/a² + y²/b² = 1. Wait, let me confirm that. When rotating around the y-axis (the minor axis), each point (x, y) on the original ellipse will trace out a circle in the x-z plane with radius x. So the equation in 3D should indeed be (x² + z²)/a² + y²/b² = 1. That seems right.Now, slicing this spheroid with a plane parallel to the axis of rotation (the y-axis). Let me visualize this. A plane parallel to the y-axis could be at some constant z-value, but wait, if the plane is parallel to the axis of rotation, which is the y-axis, then the plane's orientation should be such that it's not necessarily perpendicular to any particular axis. Wait, maybe I need to think about the general equation of a plane parallel to the y-axis.Planes parallel to the y-axis can be expressed in different ways. For example, a plane could be parallel to both the y-axis and another direction. Alternatively, a plane might have a normal vector that is perpendicular to the y-axis. Let me recall that the general equation of a plane is Ax + By + Cz + D = 0. For the plane to be parallel to the y-axis, its normal vector (A, B, C) should be perpendicular to the direction vector of the y-axis, which is (0,1,0). So the dot product of the normal vector and (0,1,0) should be zero. That gives B = 0. Therefore, the equation of the plane is Ax + Cz + D = 0. So such planes are vertical planes that can be tilted in the x-z directions but don't vary in the y-direction.But the problem says "slicing this spheroid with a plane parallel to the axis of rotation". Since the axis of rotation is the y-axis, the plane is parallel to the y-axis. So any plane that's parallel to the y-axis can be represented as Ax + Cz + D = 0. However, depending on the specific plane, the cross-section will vary. The problem states that slicing with such a plane results in an ellipse. So I need to show that the intersection of the oblate spheroid (x² + z²)/a² + y²/b² = 1 with any plane Ax + Cz + D = 0 is an ellipse.Alternatively, maybe it's easier to choose a specific coordinate system where the plane is simpler. For instance, since the spheroid is symmetric around the y-axis, we can rotate the coordinate system around the y-axis such that the plane equation simplifies. Let me think. If the plane is Ax + Cz + D = 0, perhaps we can rotate the coordinate system around the y-axis so that the plane becomes, say, x' = k, where k is some constant. Because any plane parallel to the y-axis can be made into a coordinate plane by rotation. Is that right?Alternatively, maybe it's simpler to parametrize the intersection. Let me try to parametrize the intersection curve. The equation of the spheroid is (x² + z²)/a² + y²/b² = 1. The plane is Ax + Cz + D = 0. To find their intersection, we can solve these two equations together.From the plane equation, we can express one variable in terms of the others. For example, solve for z: z = (-Ax - D)/C, assuming C ≠ 0. If C = 0, then the plane equation is Ax + D = 0, which is x = -D/A, a constant. Then substituting into the spheroid equation, we get ((-D/A)² + z²)/a² + y²/b² = 1. But if x is fixed, then z and y can vary. Let's see. If x is fixed, then (z²)/a² + y²/b² = 1 - (D²)/(A² a²). The left side is the equation of an ellipse in the z-y plane, provided that 1 - (D²)/(A² a²) is positive. So that cross-section is indeed an ellipse. Similarly, if the plane is of the form x = constant, then the cross-section is an ellipse.But the problem states the plane is parallel to the axis of rotation, which is the y-axis. So planes parallel to the y-axis can be of two types: either vertical planes like x = constant or z = constant, or tilted planes like Ax + Cz + D = 0. Wait, actually, x = constant and z = constant are special cases of Ax + Cz + D = 0 with either A or C being zero.So in the case where the plane is x = constant, we get an ellipse as shown above. Similarly, if the plane is z = constant, substituting into the spheroid equation gives (x² + (constant)^2)/a² + y²/b² = 1, which is also an ellipse in the x-y plane.But what if the plane is tilted, like x + z = constant? Let's check that. Let's take a plane like x + z = k. Then substituting z = k - x into the spheroid equation:(x² + (k - x)^2)/a² + y²/b² = 1.Expanding the numerator: x² + k² - 2k x + x² = 2x² - 2k x + k².So the equation becomes (2x² - 2k x + k²)/a² + y²/b² = 1.Let's rearrange terms:2x²/a² - 2k x/a² + k²/a² + y²/b² = 1.This is a quadratic equation in x and y. To check if this is an ellipse, we need to see if the quadratic terms form a positive-definite matrix and that the equation represents a closed curve.The quadratic terms are (2/a²)x² + (1/b²)y². Since a and b are positive, these coefficients are positive. The cross term here is only in x² and y²; there's no xy term. However, there is a linear term in x: -2k x/a². So this is an ellipse, but possibly translated. To confirm, we can complete the square for the x terms.Let me rewrite the equation:(2x² - 2k x)/a² + y²/b² = 1 - k²/a².Factor out 2/a² from the x terms:2/a² (x² - k x) + y²/b² = 1 - k²/a².Complete the square for x:x² - k x = x² - kx + (k²/4) - (k²/4) = (x - k/2)^2 - k²/4.So substituting back:2/a² [(x - k/2)^2 - k²/4] + y²/b² = 1 - k²/a².Multiply out the 2/a²:2/a² (x - k/2)^2 - 2k²/(4a²) + y²/b² = 1 - k²/a².Simplify the constants:2/a² (x - k/2)^2 + y²/b² = 1 - k²/a² + 2k²/(4a²) = 1 - k²/a² + k²/(2a²) = 1 - k²/(2a²).So the equation becomes:2/a² (x - k/2)^2 + y²/b² = 1 - k²/(2a²).Multiply both sides by the reciprocal to get the standard form:[(x - k/2)^2]/( (a²/2)(1 - k²/(2a²)) ) + [y²]/( b²(1 - k²/(2a²)) ) = 1.This is the equation of an ellipse centered at (k/2, 0) in the x-y plane (since z was replaced by k - x, but here we're looking at the cross-section in terms of x and y). The denominators under x and y must be positive for this to be an ellipse. So we need 1 - k²/(2a²) > 0, which implies |k| < sqrt(2) a. But since the original spheroid has x² + z² ≤ a², the maximum value of x + z would be sqrt(2) a (by Cauchy-Schwarz inequality), so k must satisfy |k| ≤ sqrt(2) a, and hence the condition is satisfied for all planes that actually intersect the spheroid.Therefore, even in the case of a tilted plane like x + z = k, the cross-section is an ellipse. Thus, regardless of the orientation of the plane (as long as it's parallel to the y-axis), the intersection with the oblate spheroid results in an ellipse.Alternatively, maybe there's a more straightforward way by using parametrization or coordinate transformation. Let's think about coordinate systems. Since the plane is parallel to the y-axis, we can perform a rotation around the y-axis such that the plane becomes x' = constant in the rotated coordinate system. Because rotating around the y-axis won't affect the y-component, and the spheroid is symmetric under such rotations.Let me elaborate. Suppose we have a plane Ax + Cz + D = 0. We can rotate the coordinate system around the y-axis by an angle θ such that in the new coordinates (x', y', z'), the plane equation becomes x' = k. The rotation is given by:x = x' cos θ - z' sin θ,z = x' sin θ + z' cos θ.Substituting into the plane equation:A(x' cos θ - z' sin θ) + C(x' sin θ + z' cos θ) + D = 0.Grouping terms:x'(A cos θ + C sin θ) + z'(-A sin θ + C cos θ) + D = 0.To eliminate the z' term, we set the coefficient of z' to zero:-A sin θ + C cos θ = 0.Solving for θ:tan θ = C/A.Therefore, if we choose θ = arctan(C/A), then the plane equation becomes x'(A cos θ + C sin θ) + D = 0.Since θ is chosen such that tan θ = C/A, then cos θ = A/sqrt(A² + C²) and sin θ = C/sqrt(A² + C²). Therefore:A cos θ + C sin θ = A*(A/sqrt(A² + C²)) + C*(C/sqrt(A² + C²)) = (A² + C²)/sqrt(A² + C²) = sqrt(A² + C²).Therefore, the plane equation becomes x' * sqrt(A² + C²) + D = 0, which simplifies to x' = -D / sqrt(A² + C²). Let's denote this constant as k = -D / sqrt(A² + C²). So in the rotated coordinate system, the plane is x' = k.Now, the equation of the spheroid in the rotated coordinates is still (x'² + z'²)/a² + y'²/b² = 1, because the spheroid is symmetric under rotations around the y-axis (which is the same as the y'-axis in the rotated system). Therefore, substituting x' = k into the spheroid equation gives:(k² + z'²)/a² + y'²/b² = 1.Rearranging:z'²/a² + y'²/b² = 1 - k²/a².This is the equation of an ellipse in the z'-y' plane, provided that 1 - k²/a² > 0, which is true as long as |k| < a. Since the original plane intersects the spheroid, this condition holds. Therefore, the cross-section is an ellipse.Hence, by rotating the coordinate system, we've shown that any plane parallel to the y-axis (axis of rotation) intersects the oblate spheroid in an ellipse. Therefore, the result is proven.Another approach might be using the general theory of quadric surfaces. An oblate spheroid is a quadric surface, and the intersection of a quadric with a plane is a conic section. To determine the type of conic, we can analyze the intersection. Since the plane is not tangent to the spheroid and intersects it, the intersection should be an ellipse. But maybe this is too high-level. The problem might require a more concrete demonstration.Alternatively, parametrize the spheroid and substitute into the plane equation. The standard parametrization of an oblate spheroid is:x = a cos θ cos φ,y = b sin θ,z = a cos θ sin φ,where θ ∈ [-π/2, π/2] and φ ∈ [0, 2π). Then, substituting into the plane equation Ax + Cz + D = 0:A a cos θ cos φ + C a cos θ sin φ + D = 0.Factor out a cos θ:a cos θ (A cos φ + C sin φ) + D = 0.Solving for θ:cos θ = -D / [a (A cos φ + C sin φ)].For real solutions, the right-hand side must be between -1 and 1. Assuming the plane intersects the spheroid, such solutions exist. Then, for each φ, θ is determined. However, expressing this as a parametric equation for the intersection curve might not directly show it's an ellipse. Maybe converting back to Cartesian coordinates would help.Alternatively, think in terms of linear algebra. The general equation of the spheroid is quadratic, and the plane is linear. Their intersection is a quadratic curve, which can be an ellipse, parabola, hyperbola, etc. To determine which one it is, we can compute the discriminant. But since the spheroid is a closed surface, any intersection with a plane should be a closed curve, hence an ellipse. But this is again a high-level argument. The problem likely expects a more hands-on approach.But given that we did the coordinate rotation approach earlier, which is concrete, and showed that in the rotated system the cross-section is an ellipse, that should suffice. The key idea is that any plane parallel to the axis of rotation can be transformed via rotation around that axis into a simple coordinate plane (like x' = k), where the cross-section clearly becomes an ellipse. Since rotations preserve the shape of the ellipse (they are rigid transformations), the original intersection must also be an ellipse.Therefore, the conclusion is that slicing the oblate spheroid with any plane parallel to the axis of rotation results in an ellipse.

✅:To demonstrate that slicing an oblate spheroid with a plane parallel to its axis of rotation yields an ellipse, follow these steps:1. Equation of the Oblate Spheroid: An oblate spheroid formed by rotating the ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) around its minor axis (the (y)-axis) has the 3D equation: [ frac{x^2 + z^2}{a^2} + frac{y^2}{b^2} = 1. ]2. General Plane Parallel to the Axis: A plane parallel to the (y)-axis (axis of rotation) can be expressed as (Ax + Cz + D = 0), where (A), (C), and (D) are constants. This plane does not involve (y), allowing (y) to vary freely.3. Coordinate Rotation: Rotate the coordinate system about the (y)-axis by an angle (theta) such that the plane simplifies to (x' = k) (eliminating (z')). The rotation transforms (x) and (z) as: [ x = x'costheta - z'sintheta, quad z = x'sintheta + z'costheta. ] Choose (theta = arctanleft(frac{C}{A}right)) to align the plane with (x' = k).4. Transformed Spheroid Equation: Substituting (x' = k) into the spheroid equation gives: [ frac{k^2 + z'^2}{a^2} + frac{y'^2}{b^2} = 1 quad Rightarrow quad frac{z'^2}{a^2} + frac{y'^2}{b^2} = 1 - frac{k^2}{a^2}. ] This is the standard form of an ellipse in the (z'-y') plane, scaled by (frac{a^2}{1 - frac{k^2}{a^2}}) and (frac{b^2}{1 - frac{k^2}{a^2}}).5. Conclusion: Since the intersection in the rotated coordinates is an ellipse and rotations preserve the shape, the original intersection must also be an ellipse.Final Answer: Slicing an oblate spheroid with a plane parallel to its axis of rotation results in an ellipse, as shown by coordinate transformation and substitution. (boxed{text{The cross-section is an ellipse.}})

🔑:1. Consider the Given Ellipse and the Ellipsoid: - We start with an ellipse with semi-major axis (a) and semi-minor axis (b) such that (a > b). - This ellipse is revolved around its minor axis (vertical axis here) to generate a surface known as a "flat prolate spheroid" or "oblate spheroid".2. Transformation by Compression: - The ellipse can be visualized as being derived from a circle through a linear compression along the major axis direction by a factor of (b/a). - For any point (K) on the generating circle, the projection onto the major axis is (V), and (E) is the point on the ellipse where (VE = frac{b}{a} VK).3. Revolving the Circle: - As the circle is revolved around the minor axis, it forms a sphere, with the circle in any plane through the axis producing an ellipse in the spheroid. - Viewing the process through a vertical plane through the axis assists in understanding how the spheroid and ellipses relate.4. Analyzing the Slices of the Ellipsoid with Vertical Planes: - Any vertical plane (S) intersecting the ellipsoid will also intersect the generating sphere. - Suppose the intersection of (S) with the sphere produces a circle (k). The center of (k) lies on the equatorial plane of the ellipsoid, parallel to the rotation axis.5. Compression of the Intersection Circle: - The intersection circle (k) in the sphere must be compressed by (frac{b}{a}) vertically. - This compression transforms (k) into an ellipse in the intersecting plane (S) of the ellipsoid.6. Conclusion: - Therefore, any vertical plane slice of the ellipsoid intersects the surface in an ellipse. - We have shown that slicing an ellipsoid with a plane parallel to the rotational axis always yields an ellipse. [boxed{text{Ellipse}}]

🧠:Okay, so I need to figure out how to represent numbers using one, two, or three fours with various arithmetic operations. The example given is 64 using two fours. Let me start by understanding the problem better.First, the task is to determine which numbers can be expressed using one, two, or three fours and different arithmetic operations. The example uses two fours to make 64. I guess the goal is to see for each number, how it can be formed with the specified number of fours. But maybe the question is more about, given a number like 64, how do you form it with two fours? The user might be asking for methods to create numbers using up to three fours, but the example is specific. Let me focus on the example first: 64 with two fours.So, two fours. What operations can we use? Probably addition, subtraction, multiplication, division, exponentiation, factorials, square roots, concatenation (like using two 4s as 44), decimal points (like .4), etc. Maybe even other operations like percentages, but I should stick to the more common ones first.Starting with the example of 64. Let me think. 64 is a power of 4, actually. 4 cubed is 64, but that would be 4^3, which uses a 3. But we can only use two fours. Wait, but maybe we can use another operation. For example, if we take 4 and 4, perhaps 4 to the power of something. Let's see.Wait, 4 squared is 16, 4 cubed is 64. But how do we get the exponent 3 using another 4? Hmm. Alternatively, if we use 4 factorial, which is 24. That's 4! = 24. But 24 isn't 64. Alternatively, maybe combining operations. Let's see.Alternatively, using two fours, maybe concatenation: 44. Then 44 plus something. But 44 + 4 would be three fours. Wait, but the example is two fours. So 44 + 4 is not allowed here. Wait, the example is 64 with two fours. So maybe 4^4? But 4^4 is 256, which is way larger. Wait, that's not right. Wait, 4^3 is 64, but how to get the 3 from another 4?Alternatively, maybe using square roots. For example, sqrt(4) is 2. Then, if we have two fours, sqrt(4) * sqrt(4) is 2*2=4. Not helpful. Wait, but maybe 4^(sqrt(4)+sqrt(4)). Wait, sqrt(4) is 2, so 2+2=4, then 4^4=256. Not helpful.Wait, maybe using factorial in a different way. For example, (4 + 4)! That's 8!, which is 40320. Way too big. Not helpful.Alternatively, 4! * 4. That's 24 * 4 = 96. Not 64. Hmm.Wait, perhaps using decimal points. For example, 4.4, but that might not help. 4.4 * 4 is 17.6. Not helpful. Alternatively, 44 is a two-digit number, but 44 is 44. Not 64. Unless we use something else. Wait, 4! + 4! = 24 + 24 = 48. Still not 64.Wait, maybe exponents and multiplication. Let's see. 4 squared is 16, 16 * 4 is 64, but that's three fours: 4^2 * 4. But we need two fours. So maybe using exponents with another base.Wait, how about 4^(sqrt(4)+sqrt(4))? Wait, sqrt(4) is 2, so 2 + 2 = 4, so 4^4=256. No, not helpful. Alternatively, 4^(sqrt(4)*sqrt(4)) which is 4^(2*2)=4^4=256 again. Not helpful.Wait, maybe using a different operation. Like, 4 << 4. But that's a bitwise operation, shifting left. 4 in binary is 100, shifting left by 4 gives 1000000 which is 64. But is that allowed? The problem says "arithmetic operations", which might not include bitwise. So maybe not. Let me check the problem statement again: "using various arithmetic operations". So probably standard arithmetic operations: +, -, *, /, exponents, roots, factorials, decimals, concatenation (maybe). But bitwise operations might be stretching it.Alternatively, maybe using percentage. Like 4% as 0.04, but that seems unlikely. Let me think differently.Wait, 4 squared is 16, but if I can get 16 from two fours, then 16 * 4 is 64 with three fours. But the example says two fours. So perhaps there's another way. Wait, the example must be possible. Let me check the example again. How is 64 represented with two fours? Maybe using 4 and 4, but how?Ah! Maybe using 4^3, but where does the 3 come from? If we can get a 3 from a 4, but using only one 4. For example, sqrt(4) is 2, then sqrt(4)+sqrt(4)/sqrt(4) would be 2 + 2/2 = 3, but that's multiple fours. Wait, but we need to use two fours total. So if I use one four to make a 3, and another four to make the base, but that might not work.Wait, perhaps using factorial in a different way. For example, (4 + 4)! is 8! which is way too big. 4! is 24, which is 24. Maybe 24 + something. But with two fours, how?Wait, another thought: 4! + 4! = 48, which is half of 96. Not 64. Alternatively, 4! * (4/ something). But that would need division.Wait, perhaps using exponents with decimal points. For example, 4^(4) is 256, but 4^(3) is 64. But how to get 3 from a single 4? If we have two fours, maybe one 4 is used as 4, and the other is used to make 3. For instance, 4 - (4/4) = 3, but that uses three fours. Wait, but we only have two fours. So maybe that's not possible.Alternatively, maybe using square roots. sqrt(4) is 2. Then, 4^(sqrt(4) + sqrt(4)) would be 4^(2 + 2) = 4^4 = 256. Still not 64. Wait, but 4^(sqrt(4) * sqrt(4)) is 4^(2*2) = 4^4 = 256. Not helpful.Wait, another approach: 4 * 4 = 16. But 16 is not 64. 16 * 4 = 64, but that's three fours. Wait, but if we can use exponents, 16 is 4 squared, so 4^(sqrt(4) + sqrt(4)) again.Wait, maybe using decimal fractions. For example, 4/.4 = 10. But how does that help? 10 * 6.4 = 64, but that would need more fours.Alternatively, concatenation: 44. Then, 44 + 20 = 64, but how to get 20 from another four? Wait, but we only have two fours. So 44 + 20 is not possible here.Wait, another idea: Maybe using the factorial and division. 4! / 4 = 6. Then 6 * something. But again, two fours. Wait, 4! / 4 = 6, but that uses two fours. Then how to get 64 from 6? Doesn't seem helpful.Wait, maybe using exponents in a different way. If I take 4 to the power of (something involving 4). For example, 4^(log base 4 of 64). But log base 4 of 64 is 3, since 4^3=64. But how to get log base 4 of 64 using another 4? That seems circular.Alternatively, maybe using a radical. Like, the cube root of 64 is 4. But how to express that with two fours. Cube root of (4?) but I need to get 64.Wait, maybe it's simpler than that. If we allow concatenation of operations, maybe something like 4! + 4! + 4! + 4! = 24*4=96, but that's four fours. Wait, maybe not.Wait, here's a thought: 4! + 4! = 48. Then 48 + 16 = 64. But 16 is 4 squared. So that would be 4! + 4! + 4^2, but that's three fours. Not allowed for the example.Wait, the example says 64 with two fours. Maybe the solution is 4^4 / 4, but that's three fours. Wait, 4^4 is 256, divided by 4 is 64. But that's three fours. No.Wait, maybe using a decimal point. For instance, .4 as 0.4. Then, 4 / 0.4 = 10. But 10 is not helpful. 4 / 0.4 = 10, so 10 * 6.4 = 64, but how?Alternatively, 4^(sqrt(4)) is 4^2=16. Then 16 * 4 =64, but again three fours.Wait, maybe using a different combination. For example, (4 + sqrt(4))! That would be (4 + 2)! = 6! = 720. No, too big.Wait, what if we use 4! and then do something else. 4! is 24. 24 + 40 = 64. 40 could be 4*10. But 10 is 4 + 4 + sqrt(4), which is three fours. Not helpful.Alternatively, maybe using percentages. 4% is 0.04, but I don't see how that helps.Wait, here's another idea: If we use 4 and 4, maybe 4^3 =64, but where does the 3 come from? If we can write 3 using one 4. For example, sqrt(4) + (4/4) = 2 +1=3, but that uses two fours. So total of three fours: 4^(sqrt(4) + (4/4)) =4^3=64. But that's three fours. But the example is two fours. So this isn't the solution.Wait, perhaps the example is using a different kind of operation. Maybe concatenation and exponentiation. For example, 4 and 4 concatenated as 44, but 44 isn't 64. Wait, maybe 4! + 4 = 24 +4=28. Not helpful.Alternatively, 4! * sqrt(4) =24*2=48. Still not 64.Wait, maybe 4^sqrt(4) * sqrt(4). That's 16 * 2=32. Not helpful.Wait, maybe using decimal fractions. For example, .4, but .4 is 0.4. Then 4 / 0.4 = 10. Then 10 * 6.4 =64. But 6.4 is 16/2.5, which might need more fours.Alternatively, maybe 4! + 4! + 4! + 4 =24+24+24+4=76. Not 64.Wait, I'm stuck. Maybe the example is using a different approach. Let me think again. The user says "how can 64 be represented using two fours?" So there must be a way. Let me search my memory for common four fours puzzles. Wait, sometimes people use square roots or exponents. Wait, 4 raised to the power of (4), that's 256. 4 cubed is 64. But how to get cube with two fours. Wait, if we use the cube root, then cube root of 64 is 4. But that's the reverse.Wait, perhaps using a different notation. Like, 4! with a dot over it for repeating decimal? Not sure. Or maybe using a different base? The problem doesn't specify base 10, but I think it's assumed.Wait, here's a different idea: Using the factorial notation in a way that allows for multiple factorials. For example, 4!! is double factorial, which is 4*2=8. But 8 is not helpful. Then 8 * 8 =64, but that's two eights, but made from two fours each, which would be four fours total.Alternatively, maybe using 4! and some other function. For example, 4! =24. Then 24 + 40 =64. But 40 is 4*10, and 10 can be 4 + 4 + sqrt(4). But again, too many fours.Wait, maybe using a decimal point with concatenation. Like, 4.4, but 4.4 squared is about 19.36. Not helpful.Wait, another approach: 64 is 2^6. So if we can get 2 from a four, and 6 from another four. For example, sqrt(4)=2 and 4 + sqrt(4)=6. Then 2^(4 + sqrt(4))=2^6=64. But this uses three fours: one for the base 2 (sqrt(4)), and two for the exponent (4 + sqrt(4)). So total of three fours. But we need two.Alternatively, maybe combining operations. For example, 4 * 4 =16. Then 16 * 4 =64. But that's three fours.Wait, maybe using exponents and multiplication. 4^(sqrt(4)) * sqrt(4) =16*2=32. Not helpful.Wait, here's an idea: Using 4 and 4 with exponentiation and a decimal point. For example, 4^(4 - 0.4). But 4 - 0.4 is 3.6, and 4^3.6 is about 4^(3 + 0.6)=4^3 *4^0.6≈64 * 2.3≈147. Not 64.Alternatively, 4^(sqrt(4) + sqrt(4)) which is 4^(2+2)=256. No.Wait, I'm really stuck. Maybe the answer is something like (4 + 4) * (4 + 4) =64, but that's four fours. Or (4 * 4) * (4) =64, which is three fours. Hmm.Wait, maybe using logarithms. If we take log base 4 of 64, that's 3. So log_4(64) =3. But how to express 64 as 4^3 using two fours? Since 3 is not directly obtainable from a single four. Unless we do something like 4^(sqrt(4) + (4/4)), but that's three fours again.Wait, perhaps the solution uses a different operation that I haven't considered, like square roots stacked. For example, sqrt(sqrt(4)) is 2^(1/4). Not helpful. Wait, sqrt(4) is 2, sqrt(4) is 2. So 2 * 2=4. Not helpful.Wait, maybe using percentages. 400% is 4. So 400% of 16 is 64. But 16 is 4 squared. So 400% * 4^2=64. But that uses three fours: two for 4^2 and one for 400%.Alternatively, using 4! and division. 4! is 24. 24 / (4/16) =24 * (16/4)=24*4=96. Not 64.Wait, another idea: 4^3 =64. If we can write 3 as 4 - (4/4), which uses three fours. But then total would be four fours. Not allowed.Wait, maybe the solution is simpler. Maybe using 4 and 4 in a different way. For example, 4 * 4 * 4 =64, but that's three fours. So not.Wait, perhaps using decimal concatenation. Like, 4.4 * something. But 4.4 * 14.545… ≈64, but that's not precise.Wait, here's a possible solution: 4! + 4! + 4 + 4 =24+24+4+4=56. Not enough.Wait, what if we use exponents with a decimal? For example, 4^(sqrt(4) + sqrt(4)/sqrt(4))=4^(2 + 0.5)=4^2.5=32. No.Wait, maybe 4! * sqrt(4) + 4 =24*2 +4=52. Still not 64.Alternatively, 4! * sqrt(4) * sqrt(4)=24*2*2=96. No.Wait, maybe I'm overcomplicating. The example might be using concatenation and exponents. Like, 4^4 / 4 = 256 /4 =64. But that's three fours. Alternatively, if we use 44 and some operation. 44 + 20=64. How to get 20 from a four? 4*5=20. 5 is 4 + 4/4. So 44 + 4*(4 +4/4). But that's five fours. Not helpful.Wait, perhaps using a different base. For example, in base 8, 4*4=20 (which is 16 in decimal), but 20 in base 8 is 16. Not helpful.Wait, maybe the solution is using a combination of factorial and square roots. For example, (sqrt(4) + sqrt(4))! = (2 +2)! =4! =24. Not helpful.Wait, here's a new idea: Using the gamma function, where gamma(n) = (n-1)!. So gamma(4)=6, but that's stretching the rules. Gamma(5)=24. Not helpful.Alternatively, using 4!! which is double factorial. 4!!=4*2=8. 8*8=64. But each 8 uses a four, so two fours to make two eights, then multiply them. So 4!! *4!!=8*8=64. That uses two fours. Wait, is that valid? Because 4!! is 8, and using two fours as 4!! * 4!!. So each 4!! is one four. Therefore, 4!! * 4!! =8*8=64. So that's two fours. That might be the solution.Yes! Because double factorial of 4 is 4*2=8. So two fours, each turned into 8 via double factorial, then multiplied together. 8*8=64. So that's 4!! *4!!=64. So using two fours with double factorial and multiplication.But wait, does the problem allow double factorial? The original problem says "various arithmetic operations". Double factorial is a valid mathematical operation, though less common. So if allowed, then yes, that works.Alternatively, if double factorial isn't allowed, then maybe another approach. But given that this works, it's a possible answer.So the answer would be 4!! * 4!! =8*8=64.Another way: sqrt(4)^(4+4). Wait, sqrt(4)=2. Then 2^(8)=256. No. Wait, sqrt(4)^(4* sqrt(4))=2^(4*2)=2^8=256. Not helpful.Alternatively, (4 +4)^sqrt(4)=8^2=64. Wait, that's three fours: 4+4=8, sqrt(4)=2, 8^2=64. So three fours. Not allowed for the example.But wait, the example is two fours. So if we can write 8^2 using two fours. 8 is 4+4, and 2 is sqrt(4). So total of three fours: two for 8 and one for 2. So (4+4)^sqrt(4)=8^2=64. But three fours.Alternatively, if we can use concatenation: 44 is two fours, then 44 + something=64. But 44 +20=64. How to get 20 from a four? Maybe 4*5=20. But 5 is 4 + (4/4). So that's two fours for 5. So total fours: two for 44, two for 5, but that's four fours. Not allowed.Wait, but if we can use decimal points: 4.4* something. 4.4*14.545≈64. But not exact.Wait, another idea: Using 4! and sqrt(4). 4!=24. 24*sqrt(4)=24*2=48. 48 +16=64. But 16 is 4^2. So total fours: three.Alternatively, 4! * sqrt(4) +4^sqrt(4)=24*2 +16=48+16=64. That uses four fours. Not allowed.Hmm. So the only way I can think of is using double factorial. 4!!=8, then 8*8=64 with two fours. If that's acceptable, then that's the solution.Alternatively, maybe using a different factorial notation. Like, (4!)!!!!... but that's getting too complex.Alternatively, using 4 written in a different base. For example, 4 in base 64 is 4, but that doesn't help.Wait, another angle: The problem says "using various arithmetic operations". Maybe including exponents where the exponent is written using a four. For example, 4^4 is 256, but that's two fours. Not helpful. Wait, but if we take the square root of 4^4, sqrt(4^4)=sqrt(256)=16. Still not 64.Alternatively, cube root of 4^4 is 4^(4/3)≈6.349. Not helpful.Wait, 4^4 divided by 4 squared is 256 /16=16. Not helpful.Wait, maybe 4! multiplied by something. 24* (something)=64. That would be 64/24≈2.666. Not sure how to get that from a four.Alternatively, 4! + 4! + 4*4=24+24+16=64. That's four fours.No. The example needs two fours. So the only viable solution I can find is the double factorial one: 4!! *4!!=8*8=64.Alternatively, maybe the example uses a different kind of factorial or operation. Maybe 4! with a square root: sqrt(4!) = sqrt(24)≈4.898. Not helpful.Wait, perhaps using percentage. 4% of 1600 is 64. But 1600 is 4*400. But that's too many fours.Alternatively, 400% of 16 is 64. 400% is 4, and 16 is 4^2. So 4 *4^2=4^3=64. So that's three fours.Hmm.Wait, I think I need to accept that the solution is using double factorial. So 4!!=8, and 8*8=64 with two fours. Therefore, the answer is (4!!) * (4!!) =64.Yes, that seems to fit. Double factorial is an arithmetic operation, albeit a less common one. So if allowed, this works.Alternatively, if the problem allows concatenation of two fours as 44, and then 44 + 20=64. But 20 would need to be formed with a four, which isn't possible with one four. 4*5=20, but 5 would need another four. So 44 + (4*5)=64, which is three fours. Not allowed.So, given that, the double factorial approach is the likely solution.Therefore, the answer is 4!! * 4!! =64.

✅:To express the number 64 using two fours, we can utilize the double factorial operation along with multiplication. Here's the step-by-step solution:1. Apply Double Factorial to Each Four: The double factorial of 4 (denoted as (4!!)) is calculated as: [ 4!! = 4 times 2 = 8 ] This operation reduces each 4 to 8.2. Multiply the Results: Using the two derived values: [ 8 times 8 = 64 ] Final Expression: [boxed{4!! times 4!! = 64}] This solution creatively employs the double factorial operation to reduce each 4 to 8 and then multiplies them to achieve 64 using exactly two fours.

🔑:To express the number 64 using only two 4’s and various arithmetic operations, we can follow a step-by-step transformation using square roots and exponents. Here’s the detailed calculation:1. Start with the expression involving two 4’s:[sqrt{left(sqrt{sqrt{4}}right)^{4!}}]2. Recognize that (4!) represents the factorial of 4:[4! = 1 times 2 times 3 times 4 = 24]3. Simplify the innermost square root:[sqrt{4} = 2]4. Substitute back and raise to the power of 24:[left(sqrt{2}right)^{24}]5. Simplify the power:[left(sqrt{2}right)^{24} = 2^{12}]6. Calculate the square root of (2^{12}):[sqrt{2^{12}} = sqrt{4096}]7. Finally, evaluate the square root:[sqrt{4096} = 64]Therefore, the number 64 can indeed be expressed using two 4’s in the following manner:[sqrt{left(sqrt{sqrt{4}}right)^{4!}}]Thus we conclude:[boxed{sqrt{left(sqrt{sqrt{4}}right)^{4!}} = 64}]

🧠:Okay, let's try to tackle this problem. So, the question is: In a company of 18 people, prove that there are either four people who are mutually unfamiliar with each other or four people who are mutually familiar with each other. Hmm. This seems like a problem related to graph theory, maybe Ramsey numbers? I remember that Ramsey numbers tell us the minimum number of people needed to guarantee a certain size of a clique or an independent set. Let me recall. The classical Ramsey number R(s, t) is the smallest number n such that any graph of n nodes contains a clique of size s or an independent set of size t. So in this case, we need to show that R(4,4) ≤ 18. Wait, but isn't the actual Ramsey number for R(4,4) known? I think it's 18. So if that's the case, then this problem is exactly stating that R(4,4) = 18. Therefore, in any group of 18 people, where we represent the relationships as a graph (each person is a node, and an edge connects two people if they are familiar), there must be either a clique of size 4 (four mutual acquaintances) or an independent set of size 4 (four mutual strangers). But wait, the problem statement is phrased the other way: four mutually unfamiliar or four mutually familiar. So yes, that's exactly the same as a clique or independent set of size 4. So if R(4,4) is indeed 18, then this problem is a direct application. But maybe the question expects a proof of this fact without just citing Ramsey numbers. Because I know that exact Ramsey numbers are often hard to compute, and R(4,4) is 18, but proving that might be non-trivial.So, how would one go about proving that R(4,4) ≤ 18? I think one approach is to use induction or maybe use the fact that Ramsey numbers satisfy certain recursive inequalities. For example, I remember that R(s, t) ≤ R(s-1, t) + R(s, t-1). So, for example, R(4,4) ≤ R(3,4) + R(4,3). But since R(s,t) is symmetric, R(4,3) = R(3,4). So if I can find R(3,4), then maybe I can use that to bound R(4,4). Let me check what R(3,4) is. I believe R(3,4) is 9. Because the Ramsey number R(3,4) is the smallest n such that any graph of n nodes has a triangle (clique of size 3) or an independent set of size 4. And I think that number is 9. So then, using the recursive formula, R(4,4) ≤ R(3,4) + R(4,3) = 9 + 9 = 18. Therefore, R(4,4) ≤ 18. Which would mean that in any graph of 18 nodes, there is either a clique of size 4 or an independent set of size 4. So that's exactly what the problem is asking for. Therefore, that's the proof.But wait, perhaps they want a more detailed argument. Let me see. Maybe we can use induction or a combinatorial argument. Let's try to outline the proof step by step.First, let's recall the recursive upper bound for Ramsey numbers: R(s, t) ≤ R(s-1, t) + R(s, t-1). This is proven by considering a single vertex in the graph. This vertex has degree either at least R(s-1, t) - 1 or at most R(s, t-1) - 1. If it has degree at least R(s-1, t) - 1, then among its neighbors, we can find either a clique of size s-1 or an independent set of size t. If we find a clique of size s-1, adding the original vertex gives a clique of size s. If we find an independent set of size t, then we're done. On the other hand, if the degree is at most R(s, t-1) - 1, then the number of non-neighbors is at least n - 1 - (R(s, t-1) - 1) = n - R(s, t-1). So if n ≥ R(s-1, t) + R(s, t-1), then the non-neighbors would be at least R(s-1, t) + R(s, t-1) - R(s, t-1) = R(s-1, t). Therefore, among the non-neighbors, we can find either a clique of size s or an independent set of size t-1. If we find a clique of size s, we're done. If we find an independent set of size t-1, adding the original vertex (which is not connected to them) gives an independent set of size t.Therefore, using induction, we can build up the Ramsey numbers. For the base cases, R(1, t) = R(t, 1) = 1, since a graph with one node trivially has a clique or independent set of size 1. R(2, t) = t, because in any graph of t nodes, either there is an edge (clique of size 2) or not, but if we need to guarantee an independent set of size t, we need t nodes.But in our case, R(3,4) = 9. Let me verify that. If R(3,4) = 9, then R(4,4) ≤ 9 + 9 = 18. So that works. But how do we know R(3,4) is 9? Maybe we need to verify that as well.To compute R(3,4), we can use similar recursive inequalities. R(3,4) ≤ R(2,4) + R(3,3). R(2,4) is 4, since in any graph of 4 nodes, you either have an edge (clique of size 2) or an independent set of size 4. Wait, actually R(2,4) is 4 because the definition is the smallest n such that any graph on n nodes contains a clique of size 2 or independent set of size 4. Since in a graph of 4 nodes, if there are no edges, then the independent set is 4. If there is at least one edge, then you have a clique of size 2. So R(2,4) = 4. Similarly, R(3,3) is 6, the classic Ramsey number where any graph of 6 nodes contains a triangle or an independent set of size 3. Therefore, R(3,4) ≤ R(2,4) + R(3,3) = 4 + 6 = 10. But wait, I thought R(3,4) is 9. So this upper bound gives 10, but the actual value is 9. Therefore, the recursive inequality is not tight here.So perhaps the exact value is known through other means. Let me check my knowledge. Yes, actually, R(3,4) is indeed 9. So even though the recursive formula gives an upper bound of 10, the actual Ramsey number is 9. Therefore, we need a better argument for that. But maybe we don't need to go into the details of R(3,4) for this problem. Since we can use the recursive upper bound of R(4,4) ≤ 18 based on R(3,4) ≤ 9 + R(4,3) = 9 + 9 = 18. But if R(3,4) is actually 9, then the inequality gives R(4,4) ≤ 18, and since it's known that R(4,4) = 18, that's sufficient.Alternatively, maybe the problem expects a more direct proof, perhaps using the pigeonhole principle or considering degrees and neighbors. Let's try to think of a way to approach this without invoking Ramsey numbers directly.Suppose we have a graph of 18 vertices. Pick any vertex v. Then, v has 17 edges connected to the other vertices. Each edge is either present (familiar) or not (unfamiliar). Let's consider the number of friends (edges present) and strangers (edges absent) that v has. By the pigeonhole principle, v must have at least ⌈17/2⌉ = 9 friends or 9 strangers. So either v has at least 9 friends or at least 9 strangers.Case 1: v has at least 9 friends. Let's look at the subgraph induced by these 9 friends. If within this subgraph, there exists a clique of size 3, then together with v, we have a clique of size 4. If within this subgraph, there exists an independent set of size 4, then we have an independent set of size 4. Therefore, if the subgraph of 9 friends contains either a clique of size 3 or an independent set of size 4, we're done. But this requires that R(3,4) ≤ 9, which is true because R(3,4) = 9. Hence, in the subgraph of 9 friends, we must have either a triangle (which would give a clique of 4 with v) or an independent set of 4.Case 2: v has at least 9 strangers. Similarly, consider the subgraph induced by these 9 strangers. Since v is not connected to them, if within this subgraph there is an independent set of size 3, adding v would give an independent set of size 4. Alternatively, if the subgraph contains a clique of size 4, then we already have a clique of size 4. Therefore, we need that R(4,3) ≤ 9, which again is true because R(4,3) = R(3,4) = 9. Therefore, in the subgraph of 9 strangers, there must be either a clique of size 4 or an independent set of size 3. If there's a clique of size 4, we're done. If there's an independent set of size 3, adding v gives an independent set of size 4.Hence, in either case, we can find either a clique of size 4 or an independent set of size 4 in the original graph of 18 vertices. Therefore, this proves that R(4,4) ≤ 18. Since it's known that R(4,4) is exactly 18, this completes the proof.Alternatively, if we don't want to rely on knowing the exact value of R(3,4), we could recursively apply the same argument. For example, in the first case where we have 9 friends, we can repeat the process: take a vertex in the 9 friends, which would have at least ⌈8/2⌉ = 4 friends or 4 strangers. If it has 4 friends, we look for a triangle in that subset or an independent set of size 4. If it has 4 strangers, we look for a clique of size 4 or an independent set of size 3. But this might get too convoluted. However, using the known Ramsey numbers simplifies the proof.So, summarizing the proof:1. Take any vertex v in the graph of 18 people.2. Vertex v has at least 9 friends or 9 strangers.3. If v has 9 friends, apply R(3,4) = 9: among these 9, there's either a clique of 3 (plus v makes 4) or an independent set of 4.4. If v has 9 strangers, apply R(4,3) = 9: among these 9, there's either a clique of 4 or an independent set of 3 (plus v makes 4).5. Therefore, in either case, we find the required group of four.This completes the proof that in any 18-person graph, there's a monochromatic clique of size 4, i.e., four mutual friends or four mutual strangers.I think this is a solid line of reasoning. The key is using the recursive Ramsey number upper bounds and the known values for smaller Ramsey numbers. It might also be helpful to mention that the equality R(4,4) = 18 is established through such recursive bounds combined with specific constructions that show 17 is not sufficient, but since the problem only asks to prove for 18, we don't need to address the lower bound here.

✅:To prove that in a company of 18 people, there are either four mutually unfamiliar individuals or four mutually familiar individuals, we use concepts from Ramsey theory, specifically the Ramsey number ( R(4,4) ), which represents the smallest number ( n ) such that any graph with ( n ) vertices contains either a clique of size 4 (four mutual acquaintances) or an independent set of size 4 (four mutual strangers). Here's the step-by-step proof:--- Proof1. Ramsey Number Recursive Bound: The Ramsey number ( R(s,t) ) satisfies the inequality: [ R(s,t) leq R(s-1,t) + R(s,t-1). ] For ( R(4,4) ), this becomes: [ R(4,4) leq R(3,4) + R(4,3). ] Since ( R(s,t) = R(t,s) ), we have ( R(4,3) = R(3,4) ).2. Known Ramsey Numbers: It is established that ( R(3,4) = 9 ). This means any graph with 9 vertices contains either a triangle (clique of size 3) or an independent set of size 4. Substituting into the inequality: [ R(4,4) leq 9 + 9 = 18. ]3. Application to the Problem: Consider a graph ( G ) with 18 vertices (representing 18 people). For any vertex ( v ), it has 17 edges (relationships). By the pigeonhole principle, ( v ) must have either: - At least 9 friends (edges), or - At least 9 strangers (non-edges). Case 1: If ( v ) has 9 friends, examine the subgraph ( H ) induced by these 9 friends. Since ( R(3,4) = 9 ), ( H ) must contain either: - A clique of size 3 (mutual acquaintances). Combined with ( v ), this forms a clique of size 4. - An independent set of size 4 (mutual strangers). Case 2: If ( v ) has 9 strangers, examine the subgraph ( H' ) induced by these 9 strangers. Since ( R(4,3) = 9 ), ( H' ) must contain either: - A clique of size 4 (mutual acquaintances). - An independent set of size 3 (mutual strangers). Combined with ( v ), this forms an independent set of size 4. 4. Conclusion: In either case, ( G ) contains either a clique of size 4 or an independent set of size 4. Therefore, in any group of 18 people, there are necessarily four mutual acquaintances or four mutual strangers.--- Final AnswerBy Ramsey theory, ( R(4,4) = 18 ). Thus, in any group of 18 people, there must exist either four mutually familiar individuals or four mutually unfamiliar individuals. This follows from the recursive bound ( R(4,4) leq R(3,4) + R(4,3) = 9 + 9 = 18 ) and the pigeonhole principle applied to vertex degrees. Conclusion: (boxed{text{Such a group of four people is guaranteed to exist.}})

🔑:1. Consider a company with 18 people. We need to prove that among these 18 people, there exist four people who are either all mutual acquaintances or mutual strangers (i.e., pairwise unrelated).2. We will use the Pigeonhole Principle to assist in our proof. Specifically, for each person in the group of 18, that person can have either at least 9 acquaintances or at least 9 strangers among the other 17. Without loss of generality, denote one such person by (A). Assume person (A) has at least 9 acquaintances within the group. Let’s denote these 9 acquaintances by (B_1, B_2, ldots, B_9).3. Next, consider the acquaintanceship relations within the subset ( {B_1, B_2, ldots, B_9} ): - If any person among (B_1, B_2, ldots, B_9) has at least 6 acquaintances or at least 6 strangers among the remaining 8 people in the subset, we can apply the Erdős–Szekeres theorem (a theorem on combinatorics) to find a subset of 4 people who are either all mutual acquaintances or mutual strangers.4. Using the Erdős–Szekeres theorem, suppose there is a person (B_i) within ( {B_1, B_2, ldots, B_9} ) with at least 6 acquaintances among the others. This would mean that among any group of people larger than 5, there is always a subset of 4 people all of whom are either pairwise acquainted or pairwise strangers. Thus, within persons (A, B_1, B_2, ldots, B_9), we find four mutual acquaintances.5. If such a subset does not exist, every person within ( {B_1, B_2, ldots, B_9} ) can have at most 5 acquaintances, leaving at least 3 strangers. Then, similarly, within this relationship structure, there would still be a subset of 4 people who are mutual strangers.6. Considering both cases, whether person (A) has 9 acquaintances or 9 strangers, ensures that we can always find within the group of 18 people a group of four who are all either mutual acquaintances or mutual strangers, ensuring completeness of the solution by combinatorial counting:[ k cdot 9 / 2 = 4.5 ]Which affirms the existence of these groups by pigeonholing such required subsets either into mutual acquaintance or strangers groups. Thus concluding: Conclusion:[boxed{text{We have proved the existence of either a group of four mutual acquaintances or a group of four mutual strangers.}}]

🧠:Okay, so I need to figure out the expected number of cards that must be turned over to reach the second ace in a shuffled pack of n cards, which includes three aces. The answer is supposed to be (n+1)/2, but I need to show why that's the case. Let me start by understanding the problem better.First, let me rephrase the problem to make sure I get it. We have a deck of n cards, and exactly three of them are aces. The deck is shuffled randomly, so all permutations are equally likely. We turn over the cards one by one from the top, and we want to find the expected position of the second ace. That is, on average, where will the second ace appear when we go through the deck. The answer given is (n+1)/2, which is interesting because it's the same as the expected value of the median of two uniformly random points, but I don't know if that's related here.Hmm, maybe I can approach this using linearity of expectation? That often simplifies problems where we need to find expected values involving multiple events. Let me recall that linearity of expectation works regardless of independence, which is helpful here because the positions of the aces are dependent events.Alternatively, maybe I can model this using combinations. Since the deck is shuffled randomly, the positions of the aces are uniformly distributed among all possible combinations of three positions in the deck. So the three aces divide the deck into four segments: before the first ace, between the first and second ace, between the second and third ace, and after the third ace.Wait, that's similar to the concept of the "stars and bars" theorem in combinatorics, where we divide objects into bins. But how does that help with expectation?Let me think. If the three aces are placed randomly, the problem is similar to distributing the non-ace cards into the four segments mentioned. Each of these segments has a certain number of non-ace cards. The expected number of cards between the first and second ace, plus the position of the first ace, would give the position of the second ace. Maybe I can compute the expected length of each segment.But there are n - 3 non-ace cards, and we need to distribute them into four segments. By symmetry, each of these segments should have the same expected number of non-ace cards. Since there are four segments, each would have an expected value of (n - 3)/4 non-ace cards. Therefore, the expected position of the first ace would be 1 + (n - 3)/4, the second ace would be first ace position + 1 + (n - 3)/4, and the third ace would be second ace position + 1 + (n - 3)/4.Wait, let me check that. If each of the four segments (before first ace, between first and second, between second and third, after third) has an expected length of (n - 3)/4, then the expected position of the first ace is (n - 3)/4 + 1, because there are (n - 3)/4 non-ace cards before it. Then the second ace would be after another (n - 3)/4 non-ace cards, so the position would be [(n - 3)/4 + 1] + [(n - 3)/4 + 1]. Wait, that seems off. Let me compute step by step.First, the expected number of non-ace cards before the first ace is (n - 3)/4. Then, the first ace is at position (n - 3)/4 + 1. Then, between the first and second ace, there's another expected (n - 3)/4 non-ace cards, so the second ace would be at position [(n - 3)/4 + 1] + [(n - 3)/4 + 1] = 2*[(n - 3)/4 + 1] = (n - 3)/2 + 2 = (n - 3 + 4)/2 = (n + 1)/2. Oh, that works out! So the expected position of the second ace is indeed (n + 1)/2.Wait, but I just did this by assuming that each of the four segments has the same expected number of non-ace cards. Is that a valid assumption? Because in the case of randomly placing three aces, the four segments are indeed exchangeable, meaning that each segment is equally likely to be in any position, so their expected lengths should be the same. Therefore, each segment has expectation (n - 3)/4. Then, the position of the second ace is the first ace's position plus the number of non-ace cards between the first and second ace plus 1 (for the second ace itself). Since the first ace is expected at (n - 3)/4 + 1, and the gap between first and second is expected (n - 3)/4, adding those gives [(n - 3)/4 + 1] + [(n - 3)/4 + 1] - 1? Wait, no. Wait, maybe not. Let me break it down again.The deck is divided as follows:1. Before the first ace: expected length (n - 3)/4 non-ace cards.2. Between first and second ace: expected length (n - 3)/4 non-ace cards.3. Between second and third ace: expected length (n - 3)/4 non-ace cards.4. After the third ace: expected length (n - 3)/4 non-ace cards.Therefore, the first ace is at position (n - 3)/4 + 1. Then, the second ace is after the next segment: (n - 3)/4 non-ace cards and then the second ace. So the position of the second ace is [(n - 3)/4 + 1] + [(n - 3)/4 + 1] = 2*(n - 3)/4 + 2 = (n - 3)/2 + 2 = (n - 3 + 4)/2 = (n + 1)/2. Yes, that seems correct. So this approach gives the desired result. Therefore, the expected position is (n + 1)/2.But wait, this seems too straightforward. Let me verify with a small example where I can compute the expectation manually. Let's take n = 3. Then all cards are aces. The second ace must be at position 2. The formula (n + 1)/2 would give (3 + 1)/2 = 2, which matches. That's a trivial case.Another example: n = 4. There are three aces and one non-ace card. The possible permutations are:1. A A A X: The second ace is at position 2.2. A A X A: The second ace is at position 2.3. A X A A: The second ace is at position 3.4. X A A A: The second ace is at position 3.So there are four permutations. In two of them, the second ace is at position 2, and in two, it's at position 3. Therefore, the expectation is (2*2 + 2*3)/4 = (4 + 6)/4 = 10/4 = 2.5. According to the formula, (4 + 1)/2 = 2.5, which matches. So that works.Another test case: n = 5. Three aces and two non-aces. Let's compute the expectation.The number of possible arrangements is C(5,3) = 10.We need to list all possible positions of the aces and compute the position of the second ace each time.The possible combinations of ace positions are:1. 1,2,3: second ace at 2.2. 1,2,4: second ace at 2.3. 1,2,5: second ace at 2.4. 1,3,4: second ace at 3.5. 1,3,5: second ace at 3.6. 1,4,5: second ace at 4.7. 2,3,4: second ace at 3.8. 2,3,5: second ace at 3.9. 2,4,5: second ace at 4.10. 3,4,5: second ace at 4.Now, let's count how many times each position occurs:- Position 2: 3 times (cases 1,2,3).- Position 3: 5 times (cases 4,5,7,8).Wait, case 4: 1,3,4: second ace at 3.Case 5: 1,3,5: second ace at 3.Case 7: 2,3,4: second ace at 3.Case 8: 2,3,5: second ace at 3.That's 4 cases. Then case 6: 1,4,5: second ace at 4.Case 9: 2,4,5: second ace at 4.Case 10: 3,4,5: second ace at 4.So position 4 occurs 3 times. So total counts:- Position 2: 3.- Position 3: 4.- Position 4: 3.Wait, total is 3 + 4 + 3 = 10, which matches. So the expectation is (3*2 + 4*3 + 3*4)/10 = (6 + 12 + 12)/10 = 30/10 = 3. The formula gives (5 + 1)/2 = 3, which matches. So this works.Okay, so the formula holds in these test cases. Therefore, the approach of dividing the deck into four segments with equal expected numbers of non-ace cards seems valid. Therefore, the expected position of the second ace is (n + 1)/2.But let me see if there's another way to think about this, maybe using indicator variables or another method, to confirm.Suppose we consider the positions of the three aces. Let's denote their positions as X1, X2, X3, where X1 < X2 < X3. We need to find E[X2].Alternatively, since all possible combinations of three distinct positions are equally likely, X2 is the middle order statistic of three uniformly random positions. The expectation of the middle order statistic (the median) in this case might be (n + 1)/2. Wait, actually, for the median of three uniformly random samples without replacement from {1, 2, ..., n}, the expectation is indeed (n + 1)/2. Is that a known result?Yes, I think for the median of an odd number of uniformly random order statistics, the expectation is the middle of the range. For example, if you pick three random distinct numbers from 1 to n, the expected median is (n + 1)/2. That seems to be the case here. So in our problem, the second ace is essentially the median of the three ace positions, and thus its expectation is (n + 1)/2. Therefore, this gives another way to see the result.But maybe I need to verify why the expectation of the median of three uniformly random positions is (n + 1)/2.Alternatively, perhaps using symmetry. For any position i in the deck, the probability that it is the median of the three aces is the same as the probability that any other position j is the median. Wait, no, that's not exactly true. Positions near the ends have different probabilities. Wait, but if we consider all possible triples of positions, each position has a certain probability of being the median. However, due to the linearity and symmetry, perhaps the expectation can be calculated as the average of the positions.Wait, actually, the expectation of the median can be found by considering that for three uniformly random positions, the problem is symmetric with respect to reflection around the middle of the deck. So, for each position i, the probability that i is the median is equal to the probability that (n + 1 - i) is the median. Therefore, the expectation would be symmetric around (n + 1)/2, implying that the expectation is (n + 1)/2. That's a symmetry argument. Let me check with n=5. For n=5, the expectation was 3, which is (5 + 1)/2 = 3. For n=4, it was 2.5, which is (4 + 1)/2 = 2.5. So that seems to hold.Therefore, by symmetry, the expectation of the median of three uniformly random positions is (n + 1)/2, which gives the result.Alternatively, we can compute the expectation directly using combinatorics. Let's denote the positions of the three aces as X1 < X2 < X3. Then, the probability that X2 = k is equal to the number of ways to choose one ace before k and one ace after k, divided by the total number of ways to choose three aces.So, for a specific position k, the number of ways to have one ace before k, one at k, and one after k is (k - 1 choose 1) * (n - k choose 1). Therefore, the probability that X2 = k is [ (k - 1)(n - k) ] / (n choose 3). Then, the expectation E[X2] is the sum from k=2 to k=n-1 of k * [ (k - 1)(n - k) / (n choose 3) ].But calculating this sum might be complicated. Let me try for a small n, say n=5. Then, n choose 3 = 10.For k=2: (2-1)(5 - 2) = 1*3 = 3. So probability 3/10. Contribution to expectation: 2*3/10 = 6/10.For k=3: (3-1)(5 - 3) = 2*2=4. Probability 4/10. Contribution: 3*4/10=12/10.For k=4: (4-1)(5 -4)=3*1=3. Probability 3/10. Contribution: 4*3/10=12/10.Total expectation: (6 + 12 + 12)/10=30/10=3, which is (5 +1)/2=3. So that works. Similarly, for n=4:Sum over k=2 and k=3.For k=2: (2-1)(4 -2)=1*2=2. Probability 2/4C3=2/4=0.5. Wait, n=4, nC3=4. So for k=2: (1)(2)=2, so probability 2/4=0.5. Contribution: 2*0.5=1.For k=3: (3-1)(4 -3)=2*1=2. Probability 2/4=0.5. Contribution: 3*0.5=1.5.Total expectation: 1 +1.5=2.5=(4 +1)/2. So that works too.Therefore, the general formula for the expectation would be sum_{k=2}^{n-1} k*(k -1)(n -k)/ (n choose 3). We need to show that this sum equals (n +1)/2.But to compute this sum in general, perhaps we can use combinatorial identities or some generating functions.Alternatively, note that the expectation of the second order statistic (the median) in three uniform samples can be derived using linearity of expectation. Wait, another approach is to use the concept of indicator variables. For each position i in the deck, define an indicator variable I_i which is 1 if the second ace is at position i, and 0 otherwise. Then, the expectation E[X2] is sum_{i=1}^n i * P(I_i =1). So we need to compute P(X2 =i) for each i.But as we saw earlier, for the second ace to be at position i, there must be exactly one ace before position i, one ace at position i, and one ace after position i. Therefore, the number of such arrangements is C(i -1,1) * C(n -i,1). The total number of arrangements is C(n,3). Therefore, P(X2 =i) = [ (i -1)(n -i) ] / C(n,3). So E[X2] = sum_{i=1}^n i * [ (i -1)(n -i) / C(n,3) ].But since the second ace can't be in the first position (since you need at least two aces before), the summation starts from i=2 to i=n-1. So E[X2] = [1 / C(n,3)] * sum_{i=2}^{n-1} i(i -1)(n -i).So, we need to compute the sum S = sum_{i=2}^{n-1} i(i -1)(n -i).Let me make a substitution: let j = i -1. Then, when i=2, j=1; when i=n-1, j=n-2. So S = sum_{j=1}^{n-2} (j +1)j(n - (j +1)) = sum_{j=1}^{n-2} j(j +1)(n - j -1).Alternatively, expand the expression inside the sum:j(j +1)(n - j -1) = j(j +1)(n -1 -j).This seems complicated. Maybe expanding it:First, note that (n -1 -j) = (n -1 -j). So, let me denote k = j, then the term is k(k +1)(n -1 -k).So S = sum_{k=1}^{n-2} k(k +1)(n -1 -k).Let me expand this product:First, multiply (k)(k +1) = k^2 +k.Then, multiply by (n -1 -k):(k^2 +k)(n -1 -k) = k^2(n -1 -k) + k(n -1 -k).= k^2(n -1) - k^3 + k(n -1) -k^2.= (n -1)k^2 -k^3 + (n -1)k -k^2.Combine like terms:= -k^3 + (n -1)k^2 -k^2 + (n -1)k.= -k^3 + (n -2)k^2 + (n -1)k.Therefore, S = sum_{k=1}^{n-2} [ -k^3 + (n -2)k^2 + (n -1)k ].We can split this into three separate sums:S = - sum_{k=1}^{n-2} k^3 + (n -2) sum_{k=1}^{n-2} k^2 + (n -1) sum_{k=1}^{n-2} k.We can use known formulas for the sums of k, k^2, and k^3.Recall that:sum_{k=1}^m k = m(m +1)/2.sum_{k=1}^m k^2 = m(m +1)(2m +1)/6.sum_{k=1}^m k^3 = [m(m +1)/2]^2.Therefore, substituting m = n -2:sum_{k=1}^{n-2} k = (n -2)(n -1)/2.sum_{k=1}^{n-2} k^2 = (n -2)(n -1)(2n -3)/6.sum_{k=1}^{n-2} k^3 = [(n -2)(n -1)/2]^2.Therefore, substituting back into S:S = - [ (n -2)^2(n -1)^2 /4 ] + (n -2)[ (n -2)(n -1)(2n -3)/6 ] + (n -1)[ (n -2)(n -1)/2 ].Let me compute each term separately.First term: - [ (n -2)^2(n -1)^2 /4 ].Second term: (n -2) * [ (n -2)(n -1)(2n -3)/6 ] = (n -2)^2(n -1)(2n -3)/6.Third term: (n -1) * [ (n -2)(n -1)/2 ] = (n -2)(n -1)^2 / 2.So, S = - (n -2)^2(n -1)^2 /4 + (n -2)^2(n -1)(2n -3)/6 + (n -2)(n -1)^2 /2.Let me factor out common terms. Note that each term has (n -2)(n -1).Factor out (n -2)(n -1):S = (n -2)(n -1)[ - (n -2)(n -1)/4 + (n -2)(2n -3)/6 + (n -1)/2 ].Let me compute the expression inside the brackets:Let’s denote A = - (n -2)(n -1)/4 + (n -2)(2n -3)/6 + (n -1)/2.Compute each term:First term: - (n -2)(n -1)/4.Second term: (n -2)(2n -3)/6.Third term: (n -1)/2.To combine these, let's find a common denominator. The denominators are 4, 6, and 2. The least common multiple is 12.Convert each term to have denominator 12:First term: -3(n -2)(n -1)/12.Second term: 2(n -2)(2n -3)/12.Third term: 6(n -1)/12.Therefore, A = [ -3(n -2)(n -1) + 2(n -2)(2n -3) + 6(n -1) ] /12.Let me expand each term in the numerator:First term: -3(n -2)(n -1) = -3[(n)(n -1) -2(n -1)] = -3[n^2 -n -2n +2] = -3[n^2 -3n +2] = -3n^2 +9n -6.Second term: 2(n -2)(2n -3) = 2[2n(n -2) -3(n -2)] = 2[2n^2 -4n -3n +6] = 2[2n^2 -7n +6] =4n^2 -14n +12.Third term: 6(n -1) =6n -6.Adding them together:(-3n^2 +9n -6) + (4n^2 -14n +12) + (6n -6) =(-3n^2 +4n^2) + (9n -14n +6n) + (-6 +12 -6) =n^2 +1n +0 =n^2 +n.Therefore, A = (n^2 +n)/12.Therefore, S = (n -2)(n -1) * (n^2 +n)/12.Simplify S:S = (n -2)(n -1)n(n +1)/12.Therefore, S = n(n +1)(n -1)(n -2)/12.Therefore, going back to the expectation:E[X2] = S / C(n,3) = [n(n +1)(n -1)(n -2)/12] / [n(n -1)(n -2)/6] = [n(n +1)(n -1)(n -2)/12] * [6 / (n(n -1)(n -2))] ) = (n +1)/2.Yes! Therefore, after all that algebra, the expectation simplifies to (n +1)/2. That's consistent with the formula given. Therefore, the expected number of cards that must be turned over to reach the second ace is indeed (n +1)/2.So, summarizing, the key steps are:1. Recognize that the problem is equivalent to finding the expectation of the median of three uniformly random positions in the deck.2. Use combinatorial arguments or symmetry to deduce that the expectation is (n +1)/2.3. Alternatively, compute the expectation directly by expanding the sum and simplifying using combinatorial identities, which confirms the result.Therefore, the solution is proven.

✅:The expected number of cards to be turned over to reach the second ace in a shuffled deck of ( n ) cards (with three aces) is ( frac{n+1}{2} ). Here's the systematic explanation:1. Symmetry and Order Statistics: The three aces divide the deck into four segments. By symmetry, the expected number of non-ace cards in each segment is ( frac{n-3}{4} ). The second ace's position is the sum of: - The first segment (before the first ace): ( frac{n-3}{4} + 1 ), - The second segment (between the first and second ace): ( frac{n-3}{4} + 1 ). Adding these gives: [ left(frac{n-3}{4} + 1right) + left(frac{n-3}{4} + 1right) = frac{n-3}{2} + 2 = frac{n+1}{2}. ]2. Median of Uniformly Random Positions: The second ace corresponds to the median of three uniformly random positions. For three uniformly distributed positions, the expected median is the midpoint of the deck, which is ( frac{n+1}{2} ).3. Direct Combinatorial Calculation: Let ( X_2 ) be the position of the second ace. The probability ( P(X_2 = k) ) is proportional to the number of ways to place one ace before ( k ), one at ( k ), and one after ( k ): [ P(X_2 = k) = frac{(k-1)(n-k)}{binom{n}{3}}. ] Summing ( k cdot P(X_2 = k) ) over valid ( k ) and simplifying yields: [ E[X_2] = frac{sum_{k=2}^{n-1} k(k-1)(n-k)}{binom{n}{3}} = frac{n+1}{2}. ]Conclusion: Using symmetry, order statistics, and combinatorial summation, the expected position of the second ace is ( boxed{frac{n+1}{2}} ).

🔑:1. Define Arrangements and Reflective Pairs: For a given sequence A of n cards, including three aces, we define the reflective arrangement A' as the sequence where the position of each card in A is mirrored about the middle of the pack. That is, if a card is in position m in A, it will be in position (n+1-m) in A'. 2. Pairing Arrangements: Every possible arrangement of the cards can be paired with its reflective counterpart. Thus, we can group all possible sequence arrangements into pairs (A, A').3. Expected Position of the Second Ace in a Pair: Consider the position of the second ace in arrangement A and denote this position by m. In the reflective arrangement A', the second ace will appear in position (n+1-m).4. Average Position Calculation: For each pair (A, A'), the positions of the second ace would be m in A and (n+1-m) in A'. The average of these positions in each pair is: [ frac{m + (n+1-m)}{2} = frac{n+1}{2} ]5. Expected Value: Since each possible arrangement of the cards has been considered symmetrically, the expected position of the second ace, considering all arrangements, will be the same as the average position calculated above. Thus, the expected number of cards that must be turned over to reach the second ace is: [ mathbb{E}[text{Position of second ace}] = frac{n+1}{2} ] Conclusion:Hence, the expected number of cards that must be turned over to reach the second ace in a well-shuffled pack of n cards, including three aces, is:[boxed{frac{n+1}{2}}]