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WolframAlpha provides the solution as $$ \sum_{x=1}^{N} \frac{(T+x)!}{x!}= \frac{(N+1)(N+T+1)!-(N+1)!(T+1)!}{(T+1)(N+1)!}$$ but I am unsure as to how this is obtained.

For some background, the summation on the left was obtained from a series like this... $$ (1 + 1) \cdot (1 + 2) \cdot (1 + 3) \cdot \ldots (1 + T) +\\ (2 + 1) \cdot (2 + 2) \cdot (2 + 3) \cdot \ldots (2 + T) +\\ (3 + 1) \cdot (3 + 2) \cdot (3 + 3) \cdot \ldots (3 + T) +\\ \ldots + \\ (N + 1) \cdot (N + 2) \cdot (N + 3) \cdot \ldots (N + T) $$

If it is any help.

xrisk
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  • related question: https://math.stackexchange.com/questions/2798964/formula-for-summation-series/2799027#2799027 – farruhota May 29 '18 at 10:36

1 Answers1

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The identity is equivalent to $$\sum_{x=0}^{N} \binom{T+x}{T}=1+\frac{1}{T!}\sum_{x=1}^{N} \frac{(T+x)!}{x!} \\= 1+\frac{(N+1)(N+T+1)!-(N+1)!(T+1)!}{(T+1)!(N+1)!} =\binom{N+T+1}{T+1}. $$ Now take a look at Sum of binomial coefficients $\sum_{k=0}^{n}\binom{k+m}{m}$ where $n=N$, $m=T$.

Robert Z
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