1²+2²+3²+…+（n-1)²=n(n-1)(2n-1)⼀6

这是一个平方和公式，求证方法很多，下面这个是最基本的：

(n+1)^3-n^3=3n^2+3n+1,
n^3-(n-1)^3=3(n-1)^2+3(n-1)+1
..............................
3^3-2^3=3*(2^2)+3*2+1
2^3-1^3=3*(1^2)+3*1+1.
把这n个等式两端分别相加，得：
(n+1)^3-1=3(1^2+2^2+3^2+....+n^2)+3(1+2+3+...+n)+n,
由于1+2+3+...+n=(n+1)n/2,
代入上式得：
n^3+3n^2+3n=3(1^2+2^2+3^2+....+n^2)+3(n+1)n/2+n
整理后得：
1^2+2^2+3^2+....+n^2=n(n+1)(2n+1)/6

By M.I.
n=1
LS = 0^2 =0
RS =0
n=1 is true
Assum n=k is true
ie 1²+2²+3²+…+（k-1)²=k(k-1)(2k-1)/6
for n=k+1
LS =1²+2²+3²+…+（k-1)²+k²
= k(k-1)(2k-1)/6 +k²
= k{ [(k-1)(2k-1) +6k] /6}
= k{ (2k^2+3k+1)/6}
= k(2k+1)(k+1)/6
= (k+1)(k+1-1)(2(k+1)-1)/6 = RS
By principle of MI, it is true for all n