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……