Let a₁=1 and for n 1, a_ n+1 = 1 2 a_ n + n²-2 n-1 n²(n+1)² . Then | _ n=1 ^ (a_ n - 2 n² ) | is equal to _ _ _ _ .

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Correct answer: 0

Let b_n = a_n - 2 n^2 . b_ n+1 = 1 2 a_n + n^2-2n-1 n^2(n+1)^2 - 2 (n+1)^2 . Simplifying: b_ n+1 = 1 2 b_n + (n+1)^2 + n^2 - 2n - 1 - 2n^2 n^2(n+1)^2 = 1 2 b_n . So b_n is GP with ratio 1 2 . b₁ = 1 - 2 = -1 , so b_n = - ( 1 2 )^ n-1 . _ n=1 ^ b_n = - 1 1- 1 2 = -2 . | _ n=1 ^ (a_n - 2 n^2 ) | = 2 .

Let a₁=1 and for n 1, a_ n+1 = 1 2 a_ n + n²-2 n-1 n²(n+1)² . Then | _ n=1 ^ (a_ n - 2 n² ) | is equal to _ _ _ _ .

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