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$$ f(x) = \sum_{n\ge 0}\binom{n+k}{2k}x^n $$

$k$ is a constant integer.

First I want to find the relation between $\binom{n+k}{2k}$'s, but failed, then I change $f(x)$ to $f(k,x)$, want to figure the relations of $f(k,x)$, but still cannot find it.

Thanks!

Michael Albanese
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delta
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  • Just a thought: $n\geq k$. Or else ${n+k\choose 2k}$ won't be defined. So what you're looking at is ${2k\choose 2k}x^{k} + {2k+1\choose 2k}x^{k+1}+\dots$ –  Jul 19 '13 at 13:33
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    There is an answer posted. But better would be some help. Can you do the case $k=0$? Can you do the case $k=1$? Do these cases give you some ideas for a more general case? – GEdgar Jul 19 '13 at 13:41

1 Answers1

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$$f(x) = \sum_{n\ge 0}\binom{n+k}{2k}x^n= x^k\sum_{n\ge 0}\binom{2k+n}{2k}x^n=\frac{x^k}{(1-x)^{2k+1}}.$$ Take a look at this wiki article.

TZakrevskiy
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  • +1, but always better, when changing variables, to change the variable name, when you can. I'd write $\sum_{m\geq 0}\dots$ for the right side of the substitution. – Thomas Andrews Jul 19 '13 at 13:37
  • Thx! I learn a new equation $\sum_{n\ge 0}\binom{n}{k}x^n=x^k/(1-x)^{k+1}$ now. – delta Jul 19 '13 at 14:03