answer:Answer: 10656

answer:To solve this problem, let's analyze each option step by step:Option A: E(xi -2) = E(xi)The expectation of a random variable xi minus a constant is given by:E(xi - 2) = E(xi) - 2Therefore, option A is incorrect because subtracting a constant from a random variable affects its expectation.Option B: Var(xi -2) = Var(xi)The variance of a random variable is not affected by subtracting a constant. This is because variance measures the spread of the random variable's values around the mean, and subtracting a constant shifts all values equally, leaving their spread unchanged. Thus:Var(xi - 2) = Var(xi)Therefore, option B is correct.Option C: E(xi^2) geq [E(xi)]^2This statement is related to the variance of a random variable xi, where the variance is defined as:Var(xi) = E(xi^2) - [E(xi)]^2Since variance is always non-negative, we have:E(xi^2) - [E(xi)]^2 geq 0Which implies:E(xi^2) geq [E(xi)]^2Therefore, option C is correct.Option D: Var[(xi -2)^2] = Var(xi^2)To verify this, let's calculate both variances with a=b=c=frac{1}{3}:- For Var[(xi-2)^2], we calculate the expectation and variance as follows:E[(xi-2)^2] = (1-2)^2 times frac{1}{3} + (2-2)^2 times frac{1}{3} + (3-2)^2 times frac{1}{3} = frac{2}{3}Var[(xi-2)^2] = (1-frac{2}{3})^2 times frac{1}{3} + (0-frac{2}{3})^2 times frac{1}{3} + (1-frac{2}{3})^2 times frac{1}{3} = frac{2}{9}- For Var(xi^2), we calculate:E(xi^2) = 1 times frac{1}{3} + 4 times frac{1}{3} + 9 times frac{1}{3} = frac{14}{3}Var(xi^2) = (1-frac{14}{3})^2 times frac{1}{3} + (4-frac{14}{3})^2 times frac{1}{3} + (9-frac{14}{3})^2 times frac{1}{3} = frac{294}{27} = frac{98}{9}Since Var[(xi-2)^2] = frac{2}{9} and Var(xi^2) = frac{98}{9}, and these two values are not equal, option D is incorrect.Therefore, the correct options are B and C. Hence, the final answer is encapsulated as:boxed{B text{ and } C}

answer:Let b i = i/3 i , then a ij = b i 2 b j (b i + b j ). Let the sum be k. Then k = ∑ a ij = 1/2 ∑ (a ij + a ji ) = 1/2 ∑ b i b j . We have the familiar 1 + 2x + 3x 2 + ... = 1/(1 - x) 2 , so x + 2x 2 + 3x 3 + ... = x/(1 - x) 2 , and hence ∑ b i = 1/3 (2/3) -2 = 3/4. So k = 1/2 (3/4) 2 = 9/32. 60th Putnam 1999 © John Scholes [email protected] 14 Dec 1999

answer:1. Define the function and its properties: Let ( f(a) = a^x ) for some fixed ( x ) and ( a > 2 ). We need to analyze the convexity or concavity of ( f(a) ).2. Second derivative of ( f(a) ): [ f(a) = a^x ] [ f'(a) = x a^{x-1} ] [ f''(a) = x (x-1) a^{x-2} ] The sign of ( f''(a) ) determines the convexity or concavity of ( f(a) ): - If ( x ge 1 ), then ( f''(a) ge 0 ), so ( f(a) ) is convex. - If ( 0 < x < 1 ), then ( f''(a) < 0 ), so ( f(a) ) is concave. - If ( x le 0 ), then ( f''(a) ge 0 ), so ( f(a) ) is convex.3. Applying Karamata's inequality: For ( x ge 1 ), since ( f(a) ) is convex, Karamata's inequality gives: [ f(9) + f(7) + f(3) + f(3) ge f(6) + f(6) + f(5) + f(5) ] For equality to hold, we need ( f''(a) = 0 ), which implies ( x = 1 ).4. Analyzing ( 0 < x < 1 ): For ( 0 < x < 1 ), ( f(a) ) is concave, so Karamata's inequality holds in the opposite direction: [ f(9) + f(7) + f(3) + f(3) le f(6) + f(6) + f(5) + f(5) ] For equality, we still need ( f''(a) = 0 ), but there are no ( x ) in ( 0 < x < 1 ) that make ( f''(a) = 0 ).5. Analyzing ( x le 0 ): For ( x le 0 ), ( f(a) ) is convex again, and the same analysis gives ( x = 0 ) as a solution.6. Verification of solutions: - For ( x = 1 ): [ 2 (5^1 + 6^1 - 3^1) = 7^1 + 9^1 ] [ 2 (5 + 6 - 3) = 7 + 9 ] [ 2 cdot 8 = 16 ] [ 16 = 16 ] This is true, so ( x = 1 ) is a solution. - For ( x = 0 ): [ 2 (5^0 + 6^0 - 3^0) = 7^0 + 9^0 ] [ 2 (1 + 1 - 1) = 1 + 1 ] [ 2 cdot 1 = 2 ] [ 2 = 2 ] This is true, so ( x = 0 ) is a solution.Thus, the solutions are ( x = 0 ) and ( x = 1 ).The final answer is (boxed{x = 0 text{ or } 1})

answer:begin{array}{l}text { Solution: Let } k=n^{2}+(n+1)^{2} . text { By }left(a^{2}+b^{2}right)^{2}=left(a^{2}-b^{2}right)^{2}+(2 a b)^{2}, text { we have } k^{2}=(2 n+1)^{2}+[2 n(n+1)]^{2}, k^{4}=left(4 n^{4}+8 n^{3}-4 n-1right)^{2}+ quadleft(8 n^{3}+12 n^{2}+4 nright)^{2} . text { Let }left|left(4 n^{4}+8 n^{3}-4 n-1right)-left(8 n^{3}+12 n^{2}+4 nright)right| =1 text {, i.e., }| 4 n^{4}-12 n^{2}-8 n-1 mid=1 . text { If } 4 n^{4}-12 n^{2}-8 n-1=1 text {, then } 4 n^{4}-12 n^{2}-8 n=2,end{array}The left side is divisible by 4, but the right side is not, which is a contradiction;If 4 n^{4}-12 n^{2}-8 n-1=-1, then4 n(n+1)^{2}(n-2)=0 text {, }Thus, n=-1,0,2.When n=-1 or 0, k=1, which contradicts the given condition;When n=2, k=13, in this case we have13=2^{2}+3^{2}, 13^{4}=119^{2}+120^{2} text {, }k is unique.This method uses the formula from k to k^{4},left(a^{2}+b^{2}right)^{2}=left(a^{2}-b^{2}right)^{2}+(2 a b)^{2} text {, }We can also use the solution formula for Pythagorean equations to prove from k^{4} to k, but the process is more complicated and will not be detailed here.

answer:1. Base Case Verification: We start by verifying the base case for ( n = 1 ) and ( n = 2 ). For ( n = 1 ): [ x^2 - f_1 x - f_0 = x^2 - x - 1 ] This is indeed equal to ( (x^2 - x - 1)(f_0) ) since ( f_0 = 1 ). For ( n = 2 ): [ x^3 - f_2 x - f_1 = x^3 - 2x - 1 ] We can factorize this as: [ x^3 - 2x - 1 = (x^2 - x - 1)(x + 1) ] This is true since ( f_2 = 2 ) and ( f_1 = 1 ).2. Inductive Step: Assume the statement is true for ( n = k ): [ x^{k+1} - f_k x - f_{k-1} = (x^2 - x - 1)(f_0 x^{k-1} + f_1 x^{k-2} + cdots + f_{k-1}) ] We need to show it holds for ( n = k + 1 ): [ x^{k+2} - f_{k+1} x - f_k ] Using the inductive hypothesis: [ (x^2 - x - 1)(f_0 x^k + f_1 x^{k-1} + cdots + f_k) = (x^2 - x - 1)(x(f_0 x^{k-1} + f_1 x^{k-2} + cdots + f_{k-1}) + f_k) ] Simplifying the right-hand side: [ = x(x^{k+1} - f_k x - f_{k-1}) + f_k (x^2 - x - 1) ] [ = x^{k+2} - f_{k+1} x - f_k ] This completes the inductive step.3. Solving the Given Equation: Given: [ x^{2024} = f_{2023} x + f_{2022} ] Using the proven formula: [ x^{2024} - f_{2023} x - f_{2022} = (x^2 - x - 1)(f_0 x^{2022} + f_1 x^{2021} + cdots + f_{2022}) ] For the equation to hold, the right-hand side must be zero: [ (x^2 - x - 1)(f_0 x^{2022} + f_1 x^{2021} + cdots + f_{2022}) = 0 ] Since ( f_0 x^{2022} + f_1 x^{2021} + cdots + f_{2022} neq 0 ) for all ( x ), we must have: [ x^2 - x - 1 = 0 ] Solving this quadratic equation: [ x = frac{1 pm sqrt{5}}{2} ]The final answer is ( boxed{frac{1 pm sqrt{5}}{2}} ).