1^2 + 2^2+3^2 + ... + n^2=n (n+1) (2n+1) / 6

Question

Алгебра

Стахий

14 октября, 05:56

1^2 + 2^2+3^2 + ... + n^2=n (n+1) (2n+1) / 6

+4

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Нинуша · Answer 1

An=6^n (n^2-1) / n! An+1=6^n+1 ((n+1) ^2-1) / (n+1) !

lim n->∞ An+1/An = 6^n+1 ((n+1) ^2-1) / (n+1) !*n!/6^n (n^2-1) =

lim n->∞ 6^n6 ((n+1) ^2-1) 1*2*3 ... * n/1*2*3 ... * n (n+1) * 6 * (n^2-1) =

6lim n->∞ (n+1) ^2-1 / (n+1) (n^2-1) = 6lim n->∞n^2+2n+1-1 / (n+1) (n^2-1) =

6lim n->∞ n (n+2) / (n+1) (n^2-1)