证明:(1)当n=1时,命题显然成立.
(2)假设当n=k(k≥1,k为整数) 时成立,即
1*k+2*(k-1)+……+(k-1)*2+k*1=(1/6)k(k+1)(k+2)①
当n=k+1时
左式=1*(k+1)+2*k+……k*2+(k+1)*1②
②式减去①式的左式
1*(k+1)+2*k+3(k-1)+……+(k-1)*3 +k*2+(k+1)*1
1*k+2(k-1)+……+(k-2)*3 +(k-1)*2+k*1
上下两式相减
=(k+1)+[k+(k-1)+……+3+2+1]=(k+1)+(1/2)k(k+1)=(1/2)(k+1)(k+2)
∴n=k+1时
右式=(1/6)k(k+1)(k+2)+(1/2)(k+1)(k+2)=(1/6)(k+1)(k+2)(k+3)
∴n=k+1时命题成立,由数学归纳法知原命题成立