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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