0Cn+1/2*1Cn+1/3*2Cn+……+1/k*(k-1)Cn+……+1/(n+1)*nCn可以截取一小部分解释，剩下自己再想想

Question

0Cn+1/2*1Cn+1/3*2Cn+……+1/k*(k-1)Cn+……+1/(n+1)*nCn
可以截取一小部分解释，剩下自己再想想

Answer 1

0Cn+1/2*1Cn+1/3*2Cn+……+1/k*(k-1)Cn+……+1/(n+1)*nCn
=n!/n!0!+(1/2)*n!/1!(n-1)!+(1/3)*n!/2!(n-2)!+...+1/(n+1)*n!/0!n!
=1/(n+1) [ (n+1)!/n!1!+(n+1)!/2!(n-1)!+(n+1)!/3!(n-2)!+...+(n+1)!/0!(n+1)!]
=1/(n+1) [ (n+1)!/(n+1)!0!+(n+1)!/n!1!+(n+1)!/2!(n-1)!+(n+1)!/3!(n-2)!+...+(n+1)!/0!(n+1)!-1 ]
=1/(n+1) * (2^(n+1)-1)
= (2^(n+1)-1)/(n+1)
不明白的地方可以追问!