Let A_n = _ k=1 ^n (-1)^ k-1 k^2 and B_n = _ k=1 ^n (-1)^ k-1 k for any natural number n . If S = _ n=1 ²⁰⁰ A_n B_n , then the value of S is equal to

Step-by-step solution

First, let us evaluate A_n and B_n for even and odd values of n . Case 1: n is even, say n = 2m . A_ 2m = 1^2 - 2^2 + 3^2 - 4^2 + + (2m-1)^2 - (2m)^2 Grouping into pairs: A_ 2m = (1-2)(1+2) + (3-4)(3+4) + + (2m-1-2m)(2m-1+2m) A_ 2m = - (3 + 7 + 11 + + (4m-1)) This is an arithmetic progression with m terms. Its sum is m 2 [3 + (4m-1)] = m(2m+1) . Thus, A_ 2m = -m(2m+1) . Similarly, B_ 2m = (1-2) + (3-4) + + (2m-1-2m) = -1 - 1 - - 1 = -m . So, for even n , A_ 2m B_ 2m = -m(2m+1) -m = 2m+1 = n+1 . Case 2: n is odd, sa