1*n+2*(n-1)+3*(n-2)+…+n*1=1/6n(n+1)(n+2)数学归纳法证明
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1*n+2*(n-1)+3*(n-2)+…+n*1=1/6n(n+1)(n+2)数学归纳法证明
证明:当n=1时,左边=1*1=1,右边=(1/6)*1*2*3=1
左边=右边,等式成立.
假设当n=k时,等式成立.
即 1*k+2*(k-1)+……+(k-1)*2+k*1=(1/6)*k*(k+1)*(k+2)
当n=k+1时
左边=1*(k+1)+2*k+……+(k-1)*3+k*2+(k+1)*1
=[1*k+1*1]+[2*(k-1)+2]+……+[(k-1)*2+k-1]+[k*1+k]+(k+1)
(把每一项分成两项)
=[1*k+2*(k-1)+……+(k-1)*2+k*1]+[1+2+……+(k-1)+k+(k+1)]
(前一项是把n=k时成立的结果带进去,后一项是等差数列)
=(1/6)*k*(k+1)*(k+2)+(1/2)(k+1)(k+2)
=(1/6)(k+1)(k+2)(k+3)
即当n=k+1时等式也成立.
综上,原等式恒成立.
当n=1时,左=1,右=1 成立
假设n=i时成立
当n=i+1时
1*(i+1)+2*(i+1-1)+3*(i+1-2)+...+i*2+(i+1)*1
=1*i+1+2*(i-1)+2+3(i-2)+3+...+i*1+1
=1*i+2*(i-1)+3(i-2)...+i*1 + 1+2+3...+i+1
=1/6[i(i+1(i+2)]+ 1/2[(i+1)(i+2)]
=1/6[(i+1)(i+2)(i+3)]
得证
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