Question

数学归纳法证明题，急求助！！！！~

1*n+2(n-1)+3(n-2)+······+n*1=1/6*n(n+1)(n+2)

Answer 1

一、n=1时，1=1成立
二、令n=k时上式成立，则n=k+1时
1*（n+1）+2*n+3(n-1)+······+n*2+（n+1）*1
=1*n+1+2(n-1)+2+3(n-2)+3+······+n*1+n+（n+1）*1
=1*n+2(n-1)+3(n-2)+······+n*1+1+2+3+······+（n+1）
=1/6*n(n+1)(n+2)+(n+1)(n+2)/2
=1/6*（n+3）(n+1)(n+2)
由一、二得，等式成立

Answer 2

当n=1时显然成立
设当n=k时，有1*k+2(k-1)+3(k-2)+······+k*1=1/6*k(k+1)(k+2)
当n=k+1时，有
1*(k+1)+2(k+1-1)+3(k+1-2)+······+(k+1)*1
=1*k+2(k-1)+3(k-2)+······+k*1 + (1+2+3+……+k+(k+1))
=1/6*k(k+1)(k+2)+(k+1)(k+2)/2
=1/6*(k+1)(k+2)(k+3)
即n=k+1时成立
故有……