1²+2²+3²+...+n²=n(n+1)(2n+1)/6
解（n+1）³-n³=3（n+1)²-3(n+1)+1n³-(n-1)³=3n²-3n+1∴1²+2²+3²+...+n²=n(n+1)(2n+1)/6
