Question

1的二次方+2的二次方+3的二次方+...+n的二次方怎么算

Answer 1

i^2 = i(i+1)-i

= (1/3)[i(i+1)(i+2) - (i-1)i(i+1)] -(1/2)[i(i+1)- (i-1)i]

1^2+2^2+.+n^2

=∑(i:1->n) i^2

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

= (1/3)n(n+1)(n+2) -(1/2)n(n+1)

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