1.當(dāng)n=1時(shí),左邊=1,右邊=(1/6)*1*(1+1)*(1+2)=1,左邊=右邊,
所以原等式成立.
2.設(shè)當(dāng)n=k（k>=1),原等式也成立,
即1*k+2*（k-1)+3*(k-2)+...+k*1=(1/6)k(k+1)（k+2）成立.
3.當(dāng)n=k+1時(shí),原等式的左邊=1*(k+1)+2*[（k+1)-1]+3*[(k+1)-2]+...+(k+1)*1
=[1*k+1]+[2*(k-1)+2]+[3*(k-2)+3]+……+[k*1+1]
=[1*k+2*（k-1)+3*(k-2)+...+k*1]+[1+2+3+……+(k+1)]
=(1/6)k(k+1)(k+2）+(k+1)(k+2)/2,(利用了2.假設(shè))
=(1/6)(k+1)(k+2)(k+3)
而右邊=(1/6)(k+1)[(k+1)+1][(k+1)+2]=(1/6)(k+1)(k+2)(k+3),
左邊=右邊,
所以,當(dāng)n=k+1時(shí),原等式也成立.
5.綜上所述,對(duì)于任意正整數(shù)n,原等式都成立