用数学归纳法证明1乘以n+2乘以(n-1)+3(n-2)+.+n乘以1=6分之1n(n+1)(n+2)
1*n+2(n-1)+3(n-2)+……+n*1=n(n+1)(n+2)/6
用数学归纳法证明:
(1)n=1时,左边=1=右边,等式成立.
(2)假设n=k时等式成立,即
1*k+2(k-1)+3(k-2)+……+k*1=k(k+1)(k+2)/6
当n=k+1时,
1(k+1)+2k+3(k-1)+……+k*2+(k+1)*1
=(1*k+2(k-1)+3(k-2)+……+k*1)+[1+2+3+……+(k+1)]
=k(k+1)(k+2)/6+(k+1)(k+2)/2
=(k+1)(k+2)(k+3)/6,命题成立.
综合(1)(2)得
当n为任一正整数时,等式成立.
不明白上一步变成这一步咋多除了2和中间多了个k+2~=k(k+1)(k+2)/6+(k+1)(k+2)/2