1n+2(n-1)+...+(n-1)2+n1=?

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1*n+2*(n-1)+...+(n-1)*2+n*1=?1*n+2*(n-1)+...+(n-1)*2+n*1=?1*n+2*(n-1)+...+(n-1)*2+n*1=?*n+2*(n-1)+3*

1*n+2*(n-1)+...+(n-1)*2+n*1=?
1*n+2*(n-1)+...+(n-1)*2+n*1=?

1*n+2*(n-1)+...+(n-1)*2+n*1=?
*n+2*(n-1)+3*(n-2)+…+n*1=1/6n(n+1)(n+2)
数学归纳法证明
1.当n=1时,左边=1,右边=(1/6)*1*(1+1)*(1+2)=1,左边=右边,
所以原等式成立.
2.设当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.当n=k+1时,原等式的左边=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.假设)
=(1/6)(k+1)(k+2)(k+3)
而右边=(1/6)(k+1)[(k+1)+1][(k+1)+2]=(1/6)(k+1)(k+2)(k+3),
左边=右边,
所以,当n=k+1时,原等式也成立.
5.综上所述,对于任意正整数n,原等式都成立

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