Let P_k(x) = 1 + 2x + 3x^2 + + k x^ k-1 for k 1 . A sum S_n is defined as S_n = _ k=1 ^n ( P_k'(1) P_k(1) )^2 , where P_k'(x) is the derivative of P_k(x) with respect to x . If S_n can be expressed in the form n(A n^2 + B n + C) D for all n 1 , where A, B, C Z , D N , and (|A|, |B|, |C|, D) = 1 , then the value of A + B + C + D is

Step-by-step solution

First, evaluate P_k(1) : P_k(1) = 1 + 2 + 3 + + k = k(k+1) 2 Next, find the derivative P_k'(x) : P_k'(x) = 2(1) + 3(2)x + 4(3)x^2 + + k(k-1)x^ k-2 Evaluate P_k'(1) : P_k'(1) = _ r=1 ^k r(r-1) = _ r=1 ^k (r^2 - r) P_k'(1) = k(k+1)(2k+1) 6 - k(k+1) 2 = k(k+1) 2 ( 2k+1 3 - 1 ) P_k'(1) = k(k+1) 2 ( 2k-2 3 ) = k(k-1)(k+1) 3 Now, find the ratio inside the summation: P_k'(1) P_k(1) = k(k-1)(k+1) 3 k(k+1) 2 = 2(k-1) 3 Substitute this into the expression for S_n : S_n = _ k=1 ^n ( 2(k-1) 3 )^2 = 4 9 _ k=1 ^n (k-1)^2 Let j =