Question

求和1²+2²+3²+……+（n-1）² 和1²+2²+3²+……+n²

求和1²+2²+3²+……+（n-1）²
和1²+2²+3²+……+n²最好有详细过程

Answer 1

an = n^2
= n(n+1) - n
= (1/3)[ n(n+1)(n+2)- (n-1)n(n+1)] - (1/2)[ n(n+1) -(n-1)n]
Sn =a1+a2+...+an
= (1/3)n(n+1)(n+2)- (1/2)n(n+1)
=(1/6)n(n+1)(2n+1)

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

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

追问

怎么知道是这么写的？ (1/3)[ n(n+1)(n+2)- (n-1)n(n+1)] - (1/2)[ n(n+1) -(n-1)n]

追答

n(n+1) 后一个 是 (n+2）

n(n+1) 前一个 是 (n-1）
n(n+1)(n+2)- (n-1)n(n+1)

=n(n+1)[ (n+2)-(n-1)]
=3n(n+1)
=>
n(n+1) =(1/3)[n(n+1)(n+2)- (n-1)n(n+1)]