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

Question

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

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)