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I recently came across a complicated sum while working on my homework. Usually I have no issues evaluating sums, but this one has stumped me. WolframAlpha managed to find a closed form solution, but I can't seem to work out how I should go about deriving it. Here's the sum and closed form from WolframAlpha:

$$\sum_{x=1}^y\frac{\left(\frac{5}{6}\right)^{x-1}}{(x-1)!(y-x)!}=\frac{\left(\frac{11}{6}\right)^{y-1}}{(y-1)!}$$

I have tried writing out the sum term by term to look for patterns. I noticed the factorials in the denominator seem to pair up sometimes, but I haven't been able to leverage that to any use. The main way I know to handle factorials in sums is to find a power series representation that matches the sum, but this sum doesn't seem to match anything I can find.

I appreciate any hints or solutions. Ideally, a solution without any prior knowledge of the result.

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    $\begingroup$ Hint : If you multiply the sum with $(y-1)!$ , you can use the binomial theorem because the binomial coeffcient $\binom {y-1}{x-1}$ appears. $\endgroup$ – Peter Apr 28 '18 at 7:48
  • $\begingroup$ @Peter thanks! I was just forgetting that the binomial theorem exists. $\endgroup$ – superckl Apr 28 '18 at 7:58
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These should probably be enough.

Hint 1: Use the binomial theorem.

Hint 2: $\dfrac{1}{(x - 1)!(y - x)!} = \dfrac{1}{(y - 1)!}\displaystyle{y - 1 \choose x - 1}$

Hint 3: $\dfrac{11}{6} = 1 + \dfrac{5}{6}$

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  • $\begingroup$ Thanks, I was just missing the binomial theorem. $\endgroup$ – superckl Apr 28 '18 at 7:57

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