Question

怎么证1的平方一直加到n的平方等于[(n+1)*n*(2n+1)]/6

Answer 1

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