Question

1的平方+2的平方+……+n的平方的求和公式是什么？

如题~

Answer 1

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

说明；n^2=n(n+1)-n
1^2+2^2+3^2+......+n^2
=1*2-1+2*3-2+....+n(n+1)-n
=1*2+2*3+...+n(n+1)-(1+2+...+n)

由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*2+2*3+...+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+....+n(n+1)(n+2)-(n-1)n(n+1)]/3
[前后消项]
=[n(n+1)(n+2)]/3

所以1^2+2^2+3^2+......+n^2
=[n(n+1)(n+2)]/3-[n(n+1)]/2
=n(n+1)[(n+2)/3-1/2]
=n(n+1)[(2n+1)/6]
=n(n+1)(2n+1)/6

Answer 2

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