How to use the binomial theorem to calculate binomials with a negative exponent I'm having some trouble with expanding a binomial when it is a fraction. eg $(a+b)^{-n}$ where $n$ is a positive integer and $a$ and $b$ are real numbers.
I've looked at several other answers on this site and around the rest of the web, but I can't get it to make sense. From what I could figure out from the Wikipedia page on the subject, $(a+b)^n$ where $n>0$
Any help will be much appreciated.

When $n$ is a negative integer and so is $k$, then we can write
\begin{align}
\binom n k & = \frac{n!}{(n-k)! k!} \tag 1 \\[10pt]
& = \frac{n(n-1)(n-2)(n-3)\cdots(n-k+1)}{k!}. \tag 2
\end{align}
Note that although line $(1)$ assumes $n$ is a nonnegative integer, line $(2)$ does not.  Line $(2)$ makes sense if $n$ is negative or is a non-integer.
Now suppose among $a$ and $b$ the one with the smaller absolute value is $b$, i.e. we have $0 \le |b| < |a|.$  That implies $\left|\dfrac b a\right|<1,$ and we will need that to assure convergence of a series we will see below.
Now suppose we want to expand $(a+b)^m$ as a sum of powers of $a$ and powers of $b$, and $m$ is not necessarily positive and not necessarily an integer. Then
$$
(a+b)^m = a^m \left( 1 + \frac b a \right)^m = a^m \sum_{k=0}^\infty \binom m k \left(\frac b a \right)^k = \sum_{k=0}^\infty \binom m k a^{m-k} b^k.
$$
This is "Newton's binomial theorem."
Note that


*

*All of the powers to which $b$ is raised are nonnegative integers, whereas those of $a$ need not be;

*If $m$ happens to be a nonnegative integer then all of the terms in which $k>m$ are $0$, since $\dbinom m k=0$ in that case.

