I'm stuck in proving the above identity using induction. In particular, I don't know how to prove for the case n+1: $\sum_{k=0}^{n+1} k \binom{n+2}{k+1} (\frac{1}{n+1})^{k+1} = 1$ by assuming case n is true.
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I have a solution without using induction. First we calculate
$\sum_{k=0}^n kC_{n+1}^{k+1}x^{k+1}=x^2\sum_{k=0}^n kC_{n+1}^{k+1}x^{k-1}.$
Note that $\sum_{k=0}^n kC_{n+1}^{k+1}x^{k-1}=\dfrac{d}{dx}\left(\sum_{k=0}^nC_{n+1}^{k+1}x^k\right).$
We compute $\begin{aligned}\sum_{k=0}^nC_{n+1}^{k+1}x^k&=\frac{1}{x}\sum_{l=1}^{n+1}C_{n+1}^l x^l\\
&=\frac{1}{x}\left(\sum_{l=0}^{n+1}C_{n+1}^{l}x^l-1\right)\\
&=\frac{1}{x}\left[(1+x)^{n+1}-1\right].\end{aligned}$ Hence we have $\begin{aligned}\sum_{k=0}^n kC_{n+1}^{k+1}x^{k+1}&=x^2\left[\frac{(n+1)(1+x)^nx-(1+x)^{n+1}}{x^2}+\frac{1}{x^2}\right]\\
&=(n+1)(1+x)^nx-(1+x)^{n+1}+1.\end{aligned}$
Then taking $x=\dfrac{1}{n}$, we have $\sum_{k=0}^n kC_{n+1}^{k+1}\left(\frac{1}{n}\right)^{k+1}=1.$
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You meant $\sum_{k=0}^n kC_{n+1}^{k+1}x^{k+1}=x^2\sum_{k=0}^n kC_{n+1}^{k+1}x^{k-1}$ – Aman Kushwaha May 25 '22 at 08:03
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@AmanKushwaha Oh, yes. I'll edit it, thank you. – ling May 25 '22 at 08:06
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@ling very nice proof, thank you! – LCJC May 25 '22 at 11:00