怎么证1的平方一直加到n的平方等于[(n+1)*n*(2n+1)]/6

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