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I've been trying to prove the following formula (for $n > 1$ natural, $a, b$ non-zero reals), but I don't know where to start.

$$\sum_{j=1}^n \binom{n-1}{j-1} \left( \frac{a-j+1}{b-n+1} \right) \left( \frac{a}{b} \right)^{j-1} \left( \frac{b-a}{b} \right)^{n-j} = \frac{a}{b}$$

Wolfram says it's right, but so far I've been unable to give a proof. I wonder if there's a combinatorial proof to it. Any help, hint or reference will be much appreciated.

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    $\begingroup$ First, you should replace $j$ by $j+1$ and $n$ by $n+1$ to obtain $$\sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} = \frac{a}{b}.$$Have you tried induction on $n$? $\endgroup$ – Greg Martin Jul 2 '14 at 8:36
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Replace $j$ by $j+1$ and $n$ by $n+1$ rewrite as Greg Martin suggested:

\begin{align*} \sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} \tag I \end{align*}

Let $p=\dfrac{a}{b}$ and $q=1-p$ and consider:

\begin{align*} \left(q+px\right)^n &= \sum_{j=0}^{n}\binom nj p^j q^{n-j} x^j\tag 1 \\ npx\, \left(q+px\right)^{n-1} &= \sum_{j=0}^{n}j \binom nj p^j q^{n-j} x^j\tag 2 \\ \end{align*} $(1)$ is by binomial expansion, $(2)$ is by differentiating w.r.t x on both sides.

Now substituting $x=1$, $(1)$ and $(2)$ become

\begin{align*} 1 &= \sum_{j=0}^{n}\binom nj p^j q^{n-j}\tag 3 \\ np &= \sum_{j=0}^{n}j \binom nj p^j q^{n-j} \tag 4 \\ \end{align*}

From $(\mathrm{I})$,$(3)$ and $(4)$,

\begin{align*} \sum_{j=0}^n \binom nj \left( \frac{a-j}{b-n} \right) \left( \frac{a}{b} \right)^j \left( \frac{b-a}{b} \right)^{n-j} &= \frac{a}{b-n}\cdot 1 - \frac{1}{b-n}\cdot np \\ &= \frac{a\left(b-n\right)}{b-n} \\ &= \frac{a}{b} \end{align*}

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