Question

谁帮我用数学归纳法证明这道题

1*n+2*(n-1)+3*(n-2)+...+n*1=1\6n(n+1)(n+2)

Answer 1

n=1时，左边=1*1=1
右边=1/6*1*2*3=1
左边=右边，等式成立！
假设n=k时成立 (k>1)即:
1*k+2(k-1)+3(k-2)+…+(k-1)*2+k*1=(1/6)k(k+1)(k+2)
当n=k+1时；
左边
=1*(k+1)+2(k+1-1)+3(k+1-2)+…+(k+1-1)*2+(k+1)*1
=1*k+1*1+2(k-1)+2*1+…+k*1+k+(k+1)
=[1*k+2(k-1)+…+(k-1)*2+k*1]+1+2+3+…+k+(k+1)
=(1/6)k(k+1)(k+2)+1+2+3+…+k+(k+1)
=(1/6)k(k+1)(k+2)+1/2*(k+1)*(k+2)
=(1/6)(k+1)(k+2)(k+3)
=(1/6)(k+1)[(k+1)+1][(k+1)+2]
=右边
原式也成立