求数列1,4,9,16,.n^2前n项和

Question

求数列 1,4,9,16,.n^2 前n项和

Answer 1

S=1+2²+3²+.+n²
由(n+1)³=n³+3n²+3n+1
得(n+1)³-n³=3n²+3n+1
于是有：
2³-1³=3*1²+3*1+1
3³-2³=3*2²+3*2+1
.
(n+1)³-n³=3n²+3n+1
把以上各式累加得
(n+1)³-1³=3(1+2²+3²+.+n²)+3(1+2+3+.+n)+n
即(n+1)³-1³=3S+(3/2)n(n+1)+n
整理得
S=(1/6)n(n+1)(2n+1)