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I would like to show that $$\sum_{n = 0}^{\infty} \binom{n+k}{k}x^n = \frac{1}{(1-x)^{k+1}}$$ for $k \in \mathbb{N}$ and $x \in (-1, 1)$.

I have tried out a couple values for $k$ and $x \in (-1, 1)$ and the formula seems to be true.

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Induction.

Outline of proof:

Prove first for $k=0$, then induction, using that the derivative of $\frac{1}{(1-x)^{k+1}}$ is $\frac{k+1}{(1-x)^{k+2}}$.

Then use:

$$\binom{n+k}{k}\cdot \frac{n}{k+1} = \binom{n-1+(k+1)}{k+1}$$

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  • $\begingroup$ Isn't the derivative of $\frac{1}{(1-x)^{k+1}}$ $\frac{-(k+1)}{(1-x)^{k+2}}$? $\endgroup$ – user247618 Oct 21 '15 at 0:51
  • $\begingroup$ @SeojunHong Nope - the derivative of $\frac{1}{(1+x)^{k+1}}$ is $\frac{-(k+1)}{(1+x)^{k+2}}$, but you forgot the $-$ in the $1-x$. $\endgroup$ – Thomas Andrews Oct 21 '15 at 0:52
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Hint: As I wrote elsewhere, $~\displaystyle{n+k\choose k}={n+k\choose n}=(-1)^n~{-k-1\choose n}.~$ Use this, to rewrite

the expression in terms of the binomial series.

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