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I have to prove that $n\cdot(x+1)^{n-1} = \sum_{k=0}^{n}k\cdot\binom{n}{k}\cdot x^{k-1}$.

I know how to prove it by expand the binomial theorem of $(x+1)^{n}$ and then derive it.

But, I have to prove it without derive the eqation, and I dont know how to do this.

Thanks a lot.

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    $\begingroup$ That expression is incorrect. The left-hand side should be $n(1+x)^{n-1}$. $\endgroup$ – Mark Viola Mar 16 '16 at 17:56
  • $\begingroup$ Can't be true. The degree of the left is $n$ and that of the right is $n-1$. $\endgroup$ – xpaul Mar 16 '16 at 17:57
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    $\begingroup$ Hint: $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$. $\endgroup$ – André Nicolas Mar 16 '16 at 17:58
  • $\begingroup$ @xpaul edited,sorry. $\endgroup$ – Noam Mar 16 '16 at 18:01
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$$k \binom{n}{k} = \frac{n!}{(k-1)! \, (n-k)!} = \frac{n(n-1)!}{(k-1)! \, ((n-1)-(k-1))!} = n \binom{n-1}{k-1},$$ for all $k = 1, 2, \ldots, n$. Note that if $k = 0$, the LHS is zero anyway.

Therefore, $$\sum_{k=0}^n k \binom{n}{k} x^{k-1} = \sum_{k=1}^n n \binom{n-1}{k-1} x^{k-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k} x^k = n (x+1)^{n-1}.$$

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Suppose in $n$ independent trials the probability of success on each trial is $p= \dfrac x {x+1},$ so that the probability of failure on each trial is $q = \dfrac 1 {x+1}.$

Then on average the number of successes is $np = \dfrac{nx}{x+1}.$

But that average is also \begin{align} \sum_{k=0}^n k\cdot \Pr(\text{the number of successes is } k) & = \sum_{k=0}^n k \binom n k p^k q^{n-k} \\[10pt] & = \sum_{k=0}^n k \binom n k \left( \frac x {x+1}\right)^k \left( \frac 1 {x+1}\right)^{n-k}. \end{align}

Therefore $$ \sum_{k=0}^n k \binom n k \left( \frac x {x+1}\right)^k \left( \frac 1 {x+1}\right)^{n-k} = \frac {nx}{x+1}. $$

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Start with the binomial formula $$ (x+1)^n=\sum_{k=0}^n\binom nkx^k $$ and take the (formal) derivative with respect to $x$ on both sides, giving$$ n(x+1)^{n-1}=\sum_{k=0}^nk\binom nkx^{k-1}. $$

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