4+9+16+...+（n+1）^2的和?
问题描述：
4+9+16+...+（n+1）^2的和?
答
1²+2²+3²+--------+n²=n(n+1)(2n+1)/64+9+16+...+（n+1）²=1²+2²+3²+--------+(n+1)²-1=(n+1)(n+2)(2n+3)/6-1=[(n+1)(n+2)(2n+3)-6]/6=(2n³+9n²+11n)/6