We know that the series $\displaystyle \sum_{r=0}^{\infty} \binom{n}{r}x^r = (1+x)^n$ but how do we prove $\displaystyle \sum_{n=0}^{\infty} \binom{n}{r}x^n = \frac{x^r}{(1-x)^{r+1}}$?
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In this answer it is shown that $$ \binom{n}{r}=\binom{n}{n-r}=(-1)^{n-r}\binom{-r-1}{n-r} $$ Therefore, $$ \begin{align} \sum_{n=r}^\infty\binom{n}{r}x^n &=\sum_{n=r}^\infty(-1)^{n-r}\binom{-r-1}{n-r}x^n\\ &=x^r\sum_{n=0}^\infty(-1)^n\binom{-r-1}{n}x^n\\ &=\frac{x^r}{(1-x)^{r+1}} \end{align} $$
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One way is by induction on $r$. For $r=0$ it’s just
$$\sum_{n\ge 0}x^n=\frac1{1-x}\;.$$
If
$$\sum_{n\ge 0}\binom{n}rx^n=\frac{x^r}{(1-x)^{r+1}}\;,$$
then
$$\begin{align*} \sum_{n\ge 0}\binom{n}{r+1}x^n&=\sum_{n\ge 0}\left(\binom{n-1}r+\binom{n-1}{r+1}\right)x^n\\ &=\sum_{n\ge 0}\binom{n-1}rx^n+\sum_{n\ge 0}\binom{n-1}{r+1}x^n\\ &=x\sum_{n\ge 0}\binom{n}rx^n+x\sum_{n\ge 0}\binom{n}{r+1}x^n\\ &=\frac{x^{r+1}}{(1-x)^{r+1}}+x\sum_{n\ge 0}\binom{n}{r+1}x^n\;, \end{align*}$$
so
$$(1-x)\sum_{n\ge 0}\binom{n}{r+1}x^n=\frac{x^{r+1}}{(1-x)^{r+1}}\;,$$
and
$$\sum_{n\ge 0}\binom{n}{r+1}x^n=\frac{x^{r+1}}{(1-x)^{r+2}}\;.$$
Brian M. Scott
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