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.

• Use the generalized binomial theorem. This is equivalent to the Taylor series expansion, which might be a way forward that is more familiar to you. Commented Oct 16, 2016 at 21:26
• @Dr.MV Would it be something like ${\frac {1}{(1-x)^{s}}}=\sum _{k=0}^{\infty }{s+k-1 \choose k}x^{k}\equiv \sum _{k=0}^{\infty }{s+k-1 \choose s-1}x^{k}.$ then? In that case, how would one calculate the other term, e.g. the 1 in this case. Obviously that 1 won't make a difference, but say it was another variable. What would it be called? The series is infinite, so there would be no way to subtract it from the total as per usual. Commented Oct 16, 2016 at 21:39

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

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

• Thinking about it, $\lvert a\rvert \ne\lvert b\rvert$ already implies that at least one of them is non-zero, but better be safe than sorry...
– user228113
Commented Oct 16, 2016 at 21:43
• Thanks a lot :) It makes a lot more sense now. I didn't consider making the coefficient of "a" 1 Commented Oct 16, 2016 at 21:44

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.

• is $\dfrac 1 {\text{something}}$ an "expansion"? $\qquad$ Commented Oct 16, 2016 at 21:30
• @MichaelHardy Depends how you look at it. :) Commented Oct 17, 2016 at 0:45

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.

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.