<s>[INST] Negative Exponents in Binomial Theorem I'm looking at extensions of the binomial formula to negative powers. I've figured out how to do $n \choose k$ when $n < 0 $ and $k \geq 0$:
$${n \choose k} = (-1)^k {-n + k - 1 \choose k}$$
So now let's look at one case for using the binomial coefficient:
$$(1+x)^n = \sum_{k=0}^n {n \choose k}x^k$$
How do I evaluate $\sum_{k = 0}^{n}$ when $n < 0$? From searching around on the internet I think it's just an infinite series, i.e. $k$ keeps incrementing by 1 forever. But that gets me confused about
$$\begin{align*}
(a + b)^n &= a^n(1 + \frac{b}{a})^n \\ &= a^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{b}{a}\right)^k\right)\\
 &= a^n \left(1 + n \left(\frac{b}{a}\right) + \frac{(n)(n-1)}{2}\left(\frac{b}{a}\right)^2 + \cdots\right)
\end{align*}$$
and 
$$\begin{align*}
(b + a)^n &= b^n\left(1 + \frac{a}{b}\right)^n\\
&= b^n \left(\sum_{k = 0}^{n}{n \choose k}\left(\frac{a}{b}\right)^k\right)\\ 
&= b^n \left(1 + n \left(\frac{a}{b}\right) + \frac{(n)(n-1)}{2}\left(\frac{a}{b}\right)^2 + \cdots\right)
\end{align*}$$
Now the two should be equal, but in the first sum I'd never get a $b^n$ and in the second sum I'd never get a $a^n$?
 [/INST] For $a=1$, the negative binomial series simplifies to
$(x+1)^{-n}=1-nx+\frac1{2!}n(n+1)x^2-\frac1{3!}n(n+1)(n+2)x^3+....$
</s>