1+1+2+1+2+3+1+2+3+4+...+(1+2+3+4+...+n)
3个回答

1+2+3+4+...+n=n(n+1)/2=1/2n^2+n/2

故:1+(1+2)+(1+2+3)+...+(1+2+3+4+...+n)

=1/2[1^2+2^2+...+n^2]+1/2(1+2+...+n)

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

=n(n+1)(2n+1)/12+n(n+1)/4

=n(n+2)^2/6

注:1^2+2^2+...+n^2=n(n+1)(2n+1)/6