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I want to derive formula for generating function $\sum_{n=0}^{+\infty}{m+n\choose m}z^n$ because it is very often very useful for me. Unfortunately I'm stuck:

$ f(z)=\sum_{n\ge 0}{m+n\choose n}z^n= \\ \sum_{n\ge 1}{m+n-1\choose n-1}z^n+\sum_{n\ge 0}{m+n-1\choose n}z^n= \\ zf(z)+\sum_{n\ge 0}{m+n-1\choose n}z^n$

here, I'm afraid that there is a need for something more sophisticated than above trivial identity. Can you help me?

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    No, I want to derive that, assuming that I don't know that. What typos? I can correct them, but don't see.2012-07-30

2 Answers 2

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The function $\left(1 + z\right)^k$ has series expansion $\left(1 + z\right)^k = \sum_{n\ge0}\binom{k}{n}z^n$ We also have the binomial coefficient identity $\binom{n}{k} = (-1)^k\binom{-n + k - 1}{k}$ which together gives $\sum_{n\ge 0}\binom{m+n}{n}z^n = \sum_{n\ge 0}\binom{-m-1}{n}(-z)^n = \frac{1}{(1-z)^{m+1}}$ which is your required generating function.

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    Right! it seems so obvious now. I'm sorry I had to ask, I should sleep more often maybe. Thank you so much.2012-07-30
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$f_m(z).(1-z)=\sum\left(\binom{m+n}{n}-\binom{m+n-1}{n-1}\right)z^n=\sum\binom{(m-1)+n}{n}z^n=f_{m-1}(z)$

And obviously $f_0(z)=\frac{1}{1-z}$, so $f_m(z)=\frac{1}{(1-z)^{m+1}}$

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    very nice too! I also tried this way and introduced $f_m(z)$ notation but I'm ashamed now, I didn't see that $f_0(z)=\frac{1}{1-z}$ so I thought that the solution isn't explicit.2012-07-30