n(n+3)=n^2+3n,因此2*5+3*6+4*7+.+n(n+3)
=3（2+3+……+n）+（2^2+3^2+……n^2)
=3{[n(n+1)/2]-1}+[n(n+1)(2n+1)]/6 - 1
=(n^3+6n^2+5n-12)/3