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+11²+2²+...+k²+(k+1)²=k(k+1)(2k+1)...