How to use the binomial theorem to calculate binomials with a negative exponent

For $\lvert x\rvert<1$ and a real number $\alpha$, you can write $(1+x)^{\alpha}$ as the convergent series $$(1+x)^{\alpha}=\sum_{k=0}^\infty \binom{\alpha}{k} x^k$$

Were $\binom\alpha k=\frac{\alpha(\alpha-1)(\alpha-2)\cdots (\alpha-k+1)}{k!}$. For instance \begin{align}\binom{1/2}{4}&=\frac{\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac32\right)\cdot\left(-\frac52\right)}{24}=-\frac{15}{16\cdot24}\\\binom{-n}{k}&=\frac{-n\cdot(-n-1)\cdots(-n-k+1)}{k!}=(-1)^k\frac{n(n+1)\cdots(n+k-1)}{k!}=\\&=(-1)^k\frac{(n+k-1)!}{(n-1)!k!}=(-1)^k\binom{n+k-1}{k}\end{align}

Now, to use this effectively when $(a+b)^{-n}$, you need at least one between $a$ and $b$ not to be zero and $\lvert a\rvert\ne\lvert b\rvert$. Then, pick the one with the highest absolute value and factor it out. If it is $a$, this means writing all as

$$a^{-n}\left(1+\frac ba\right)^{-n}=a^{-n}\sum_{k=0}^\infty \binom{-n}{k}\frac{b^k}{a^k}=\sum_{k=0}^\infty \binom{n+k-1}{k}(-b)^ka^{-k-n}$$

Note that $(a+b)^{-n} = \frac{1}{(a+b)^n}$.

Now, apply $(a+b)^n = \sum_{i=0}^n \binom{n}{i} a^i b^{n-i}$ to calculate the denominator.

You will always get a series, and for my own part, I find the story easiest to understand when $a=1$. Then, the standard formula applies: $$ (1+b)^t=1+tb+\frac{t(t-1)}2b^2+\sum_{k=3}^\infty\binom tkb^k\,, $$ where $\binom tk=t(t-1)(t-2)\cdots (t-k+1)\big/k!$ . This is the Generalized Binomial Theorem mentioned in the comment of @Dr.MV, and it’s often proved in second-term Calculus. The formula is good and convergent whenever $|b|<1$, no matter whether $t$ is a negative integer or a rational number, or even a real number. When $a\ne1$, you’ll be mentioning $a^t$ everywhere. Note that this series expansion does not treat $a$ and $b$ symmetrically.