# binomial formula for negative integer powers

For negative integer powers, the binomial formula can be written in terms of binomial coefficients like so:

 $(1-x)^{-n}=\sum_{m=1}^{\infty}\binom{m+n-1}{n-1}x^{m}$

Proof:   We shall prove this by induction on $n$. First, note that, if $n=1$, then $\binom{m}{0}=1$, so our formula reduces to

 $(1-x)^{-1}=\sum_{m=1}^{\infty}x^{m},$

which is the formula for the sum of an infinite geometric series.

Next, suppose that the formula is valid for a certain value of $n$. Then we have

 $(1-x)^{-n-1}=(1-x)^{-1}(1-x)^{-n}=\left(\sum_{k=0}^{\infty}x^{k}\right)\left(% \sum_{m=0}^{\infty}{m+n-1\choose n-1}x^{m}\right)$

The product of sums can be rewritten as the following double sum:

 $\sum_{m=0}^{\infty}\sum_{k=0}^{m}{n+k-1\choose n-1}x^{m}$

The easiest way to see this is by rearranging the double sum as follows and adding columns

 $\begin{matrix}x^{0}\sum_{m=0}^{\infty}\binom{m+n-1}{n-1}x^{m}=&\binom{n-1}{n-1% }&+&\binom{n}{n-1}x&+&\binom{n+1}{n-1}x^{2}&+&\binom{n+2}{n-1}x^{3}&+&\binom{n% +3}{n-1}x^{4}&+&\cdots\\ x^{1}\sum_{m=0}^{\infty}\binom{m+n-1}{n-1}x^{m}=&&&\binom{n-1}{n-1}x&+&\binom{% n}{n-1}x^{2}&+&\binom{n+1}{n-1}x^{3}&+&\binom{n+2}{n-1}x^{4}&+&\cdots\\ x^{2}\sum_{m=0}^{\infty}\binom{m+n-1}{n-1}x^{m}=&&&&&\binom{n-1}{n-1}x^{2}&+&% \binom{n}{n-1}x^{3}&+&\binom{n+1}{n-1}x^{4}&+&\cdots\\ x^{3}\sum_{m=0}^{\infty}\binom{m+n-1}{n-1}x^{m}=&&&&&&&\binom{n-1}{n-1}x^{3}&+% &\binom{n}{n-1}x^{4}&+&\cdots\\ .&.&.&.&.&.&.&.&.&.&.&.\end{matrix}$

 $\sum_{k=0}^{m}\binom{n+k-1}{n-1}=\binom{m+n}{n}$
 $(1-x)^{-n-1}=\sum_{m=0}^{\infty}\binom{m+n}{n}x^{m}.$