求1*1+2*3+3*5+……+n(2n-1)的值

解：
1×1+2×3+3×5+……+n×(2n-1)
=(2×1^2-1)+(2×2^2-2)+2×3^2-3)+……＋(2n^2-n)
=2(1^2+2^2+3^2+……+n^2)-(1+2+3+……+n)
=2[n(n+1)(2n+1)/6]-(n+1)n/2
=n(n+1)(2n+1)/3-(n+1)n/2
=n(n+1)[(2n+1)/3-1/2]
=n(n+1)[(2n/3-1/6]
=n(n+1)(4n-1)/6

附1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6 的求法

利用立方差公式
n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n
=2*n^2+(n-1)^2-n

2^3-1^3=2*2^2+1^2-2
3^3-2^3=2*3^2+2^2-3
4^3-3^3=2*4^2+3^2-4
……
n^3-(n-1)^3=2*n^2+(n-1)^2-n