n=1
1的平方=1,（1+1）*1*（2+1）/6=1
所以当n=k
(k+1)*k*(2k+1)/6=1方+2方+.+k方
n=k+1也成立
1f+2f+3f+...+kf+(k+1)f
=(k+1)*k*(2k+1)/6+(k+1)f
=(k+1)*k*(2k+1)/6+6(k+1)f/6
=(k+1+1)*(k+1)*[2(k+1)+1]/6
由上可知,命题成立