數(shù)學(xué)歸納法證明：1*n+2(n-1)+3(n-2)+…+(n-1)*2+n*1=(1/6)n(n+1)(n+2)
n=1時,左邊=1*1=1
右邊=1/6*1*2*3=1
左邊=右邊,等式成立!
假設(shè)n=k時成立 (k>1)即:
1*k+2(k-1)+3(k-2)+…+(k-1)*2+k*1=(1/6)k(k+1)(k+2)
當(dāng)n=k+1時；
左邊
=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)】怎么來的?
=[1*k+2(k-1)+…+(k-1)*2+k*1]+【1+2+3+…+k+(k+1)】←為什么是一個等差數(shù)列?不是1+2+3+·······+3+2+1嗎?
=(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]
= =沒人嗎