# Binomial theorem of negative index

I have to compute fifth term of $(2x-1)^{-1}$.

I wrote it $\frac {1}{(2x-1)}$ but don't know how to proceed further. My teacher told me to look for newton's binomial theorem at wiki ,,, but I didn't understood how to compute it even when I read the article of Wikipedia of newton's theorem.

Please help me in solving it.

• Yes answers are sufficient for me. Thanks guys!! Commented Dec 7, 2016 at 11:27

## 2 Answers

Recall the geometric series $$\frac 1{1-y}= \sum_{n=0}^\infty y^n, \qquad |y| < 1.$$ Now let $y = 2x$, we get $$\frac 1{1-2x} = \sum_{n=0}^\infty 2^n x^n, \qquad |x| < \frac 12$$ From that, you can continue, I'm sure.

The answer by martini is probably enough, but to try and address what your teacher was talking about, note the following:

There is a way to extend the binomial theorem to include negative $n$. So that you could define

$(x+a)^{-n}=\sum_{k=0}^{\infty}{-n \choose k}x^ka^{-n-k}$

where you define

${-n \choose k}:=(-1)^k{n+k-1 \choose k}$.

Here we use $\dfrac{n(n-1)(n-2)\cdots(n-r+1)}{r!}$ as a definition of ${n \choose r}$.

You can read more about it in this thread.

So expanding $(2x-1)^{-1}$ (around 0) would give you:

${-1 \choose 0}(2x)^0(-1)^{-1}+{-1\choose 1}(2x)^1(-1)^{-1-1}+\cdots +{-1\choose 5}(2x)^5(-1)^{-1-5}+\cdots$

So $2^5{-1\choose 5}(-1)^{-1-5}=2^5(-1)^5{5 \choose 5}(-1)^{-6}=-2^5x^5=-32x^5$

would be your fifth term if I'm not mistaken.

I am not sure that this was what your teacher meant. In this case it was probably even a lot easier using the answer given by martini. However, it might be a nice thing to use in other cases, especially when you have an even lower negative exponent.

Hope this was at least useful for you.