Question

an=n²求数列前n项和

急求过程

Answer 1

1²+2²+3²+...+n²=n(n+1)(2n+1)/6.

可以用数学归纳法证明或裂项法证明
因为(n+1)³-n³=3n²+3n+1，
所以 n³-(n-1)³=3(n-1)²+3(n-1)+1
(n-1)³-(n-2)³=3(n-2)²+3(n-2)+1
。。。
2³-1³=3·1²+3·1+1
相加得(n+1)³-1=3(1²+2²+3²+...+n²)+3(1+2+3+...+n)+n
=3(1²+2²+3²+...+n²)+3n(n+1)/2+n
解得²+2²+3²+...+n²=n(n+1)(2n+1)/6.

Answer 2

an =n^2
= n(n+1) -n
= (1/3)[n(n+1)(n+2)-(n-1)n(n+1)] -n
∑(i:1->n) an
={∑(i:1->n) (1/3)[i(i+1)(i+2)-(i-1)i(i+1)] } -{ ∑(i:1->n) i}
= (1/3)n(n+1)(n+2) -n(n+1)/2
= (1/6)n(n+1)(2n+1)