1²+2²+3²+--------+n²=n(n+1)(2n+1)/6
4+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
平方和公式n(n+1)(2n+1)/6 即1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6
用公式：1^2+2^2+3^2+……+m^2=m(m+1)(2m+1)/6,
4+9+16+...+（n+1）^2
=(n+1)(n+2)(2n+3)/6-1.
4+9+16+...+（n+1）^2=(n+1)^3/3+(n+1)^2/2+(n+1)/6-1
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