1²+2²+...+n²=n(n+1)(2n+1)/6
用数学归纳法：
n=1时,1=1*2*3/6=1成立
假设n=k时也成立,那么k(k+1)(2k+1)/6=1²+2²+...+k²
那么n=k+1
1²+2²+...+k²+(k+1)²=k(k+1)(2k+1)/6+(k+1)²=k(k+1)(2k+1)+6(k+1)²/6
k(k+1)(2k+1)+6(k+1)²=(k+1)(2k²+k+6k+6)=(k+1)*(2k²+7k+6)=(k+1)(k+2)(2k+3)
=(k+1)((k+1)+1)(2(k+1)+1)
所以1²+2²+...+k²+(k+1)²=k(k+1)(2k+1)/6+(k+1)²=k(k+1)(2k+1)+6(k+1)²/6
=（k+1)((k+1)+1)(2(k+1)+1)/6
即n=k+1时,也成立
所以
1²+2²+...+n²=n(n+1)(2n+1)/6
过程你自己写完整,也可以推导,但是过程有点复杂...
