1+4+9+16+……+n²=n（n+1）（2n+1）/6 怎么推导的?

(n+1)³=n³+3n²+3n+1
(n+1)³-n³=3n²+3n+1
所以
2³-1³=3*1²+3*1+1
3³-2³=3*2²+3*2+1
4³-3³=3*3²+3*3+1
.
(n+1)³-n³=3n²+3n+1
将上述n项相加得
(n+1)³-1³=3*(1²+2²+3²+...+n²)+3*(1+2+3+...n)+n
(n+1)³-1³=3*(1²+2²+3²+...+n²)+3*(1+n)*n /2 +n
(n+1)³-1³-3*(1+n)*n /2 -n=3*(1²+2²+3²+...+n²)
(n+1)³-3*(1+n)*n /2 -(1+n)=3*(1²+2²+3²+...+n²)
(n+1)[(n+1)²-3n/2-1]=3*(1²+2²+3²+...+n²)
(n+1)(n² +n/2)=3*(1²+2²+3²+...+n²)
(n+1)[n(2n+1)/2]=3*(1²+2²+3²+...+n²)
1²+2²+3²+...+n²=(n+1)n(2n+1)/6=n（n+1）（2n+1）/6