# Binomial expansion, general coefficient

## Homework Statement

Find the coefficient of x^n in the expansion of each of the following functions as a series of ascending powers of x.

$\frac{1}{(1+2x)(3-x)}$

## The Attempt at a Solution

$(1+2x)^{-1} = 1 + (-1)2x + \frac{(-1)(-2)}{2!}(2x)^2 + \frac{(-1)(-2)(-3)}{3!}(2x)^3... -1<2x<1$

$= 1 - 2x + 4x^2 - 8x^3... -\frac{1}{2}<x<\frac{1}{2}$

$\frac{1}{3}(1-\frac{x}{3})^{-1} = \frac{1}{3}(1 + (-1)(-\frac{x}{3}) + \frac{(-1)(-2)}{2!}(-\frac{x}{3})^2 + \frac{(-1)(-2)(-3)}{3!}(-\frac{x}{3})^3...) -1<-\frac{x}{3}<1$

$= \frac{1}{3} + \frac{x}{9} + \frac{x^2}{27} + \frac{x^3}{81}... -3<x<3$

$\frac{1}{(1+2x)(3-x)}=\frac{1}{3}-\frac{5}{9}x+\frac{31}{27}x^2-\frac{185}{81}x^3... -\frac{1}{2}<x<\frac{1}{2}$

I can't see an obvious pattern in the numerators of the coefficients, perhaps I should be able to. So I look at the coefficient of each binomial where I see that the coefficient of $x^n$ in the expansion of $(1+2x)^{-1}$ is $(-2)^n$ and the coefficient of $x^n$ in the expansion of $(3-x)^{-1}$ is $(\frac{1}{3})^{n+1}$. So $\frac{1}{(1+2x)(3-x)}$ could also be expressed as $(\sum^{∞}_{r=0}(-2x)^r)(\sum^{∞}_{r=0}(\frac{1}{3})^{r+1}(x)^r) -\frac{1}{2}<x<\frac{1}{2}$. However as I feel I am getting tantalisingly close to a solution I realise that their multiplication complicates matters.

So I start to look at patterns in the process of multiplying the 2 expansions together and find that
$\frac{1}{(1+2x)(3-x)}=\frac{1}{3}(\sum^{∞}_{r=0} (-1)^r(2x)^r(\sum^{∞}_{k=0} (\frac{x}{3})^k)) -\frac{1}{2}<x<\frac{1}{2}$

By now I've strayed into territory outside of my text book, ie the recursive summation, this might well be meaningless as it looks like it never bottoms out of the k loop, but I hope it illustrates my thought process, however Ill conceived it might be.
I thought there might be a way to eliminate the k in the last expression to yield my general coefficient.
Any guidance on the approach to tackling this kind of problem would be much appreciated.

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pasmith
Homework Helper

## Homework Statement

Find the coefficient of x^n in the expansion of each of the following functions as a series of ascending powers of x.

$\frac{1}{(1+2x)(3-x)}$

## The Attempt at a Solution

$(1+2x)^{-1} = 1 + (-1)2x + \frac{(-1)(-2)}{2!}(2x)^2 + \frac{(-1)(-2)(-3)}{3!}(2x)^3... -1<2x<1$

$= 1 - 2x + 4x^2 - 8x^3... -\frac{1}{2}<x<\frac{1}{2}$

$\frac{1}{3}(1-\frac{x}{3})^{-1} = \frac{1}{3}(1 + (-1)(-\frac{x}{3}) + \frac{(-1)(-2)}{2!}(-\frac{x}{3})^2 + \frac{(-1)(-2)(-3)}{3!}(-\frac{x}{3})^3...) -1<-\frac{x}{3}<1$

$= \frac{1}{3} + \frac{x}{9} + \frac{x^2}{27} + \frac{x^3}{81}... -3<x<3$

$\frac{1}{(1+2x)(3-x)}=\frac{1}{3}-\frac{5}{9}x+\frac{31}{27}x^2-\frac{185}{81}x^3... -\frac{1}{2}<x<\frac{1}{2}$

I can't see an obvious pattern in the numerators of the coefficients, perhaps I should be able to. So I look at the coefficient of each binomial where I see that the coefficient of $x^n$ in the expansion of $(1+2x)^{-1}$ is $(-2)^n$ and the coefficient of $x^n$ in the expansion of $(3-x)^{-1}$ is $(\frac{1}{3})^{n+1}$. So $\frac{1}{(1+2x)(3-x)}$ could also be expressed as $(\sum^{∞}_{r=0}(-2x)^r)(\sum^{∞}_{r=0}(\frac{1}{3})^{r+1}(x)^r) -\frac{1}{2}<x<\frac{1}{2}$. However as I feel I am getting tantalisingly close to a solution I realise that their multiplication complicates matters.

So I start to look at patterns in the process of multiplying the 2 expansions together and find that
$\frac{1}{(1+2x)(3-x)}=\frac{1}{3}(\sum^{∞}_{r=0} (-1)^r(2x)^r(\sum^{∞}_{k=0} (\frac{x}{3})^k)) -\frac{1}{2}<x<\frac{1}{2}$
You can swap multiplication by $(-2x)^r$ and summation over $k$ to obtain
$$\frac{1}{(1 + 2x)(3 - x)} = \frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty \frac{(-1)^r2^r}{3^k}x^{r+k}$$
and now the change of variables to $n = r + k$ and $m = k$ where $0 \leq n \leq \infty$ and $0 \leq m \leq n$ suggests itself.

Thanks for your reply pasmith. Trying to follow your suggestion I got the coefficient of x^n as $\sum ^{∞}_{m} \frac{(-2)^{n-m}}{3^{m+1}}$ which I struggled to make sense of.
But by using your distribution of (-2x)^r I was able to visualise the whole summation as a 2 dimensional array of terms and by adding diagonal terms come up with an expression for the coefficient of every nth power of x:
$\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1} }{3^n}$
However I am still unable to express this coefficient in terms of just n and without the summation. Something tells me I'm making more of a meal of this than is necessary.

pasmith
Homework Helper
Thanks for your reply pasmith. Trying to follow your suggestion I got the coefficient of x^n as $\sum ^{∞}_{m} \frac{(-2)^{n-m}}{3^{m+1}}$ which I struggled to make sense of. But by using your distribution of (-2x)^r I was able to visualise the whole summation as a 2 dimensional array of terms and by adding diagonal terms
This is indeed the idea.

come up with an expression for the coefficient of every nth power of x:
$\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1} }{3^n}$
However I am still unable to express this coefficient in terms of just n and without the summation. Something tells me I'm making more of a meal of this than is necessary.
Your sum here is a geometric series: $$\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1}}{3^n} = \frac1{3^{n+1}} \sum_{m=0}^n (-6)^m.$$ There is a formula for summing such series.

The stricter constraint is that $|x| < \frac12$, so it makes sense to express the series in powers of $2x$ rather than $x/3$. Hence $$\frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty (-2x)^r \frac{x^k}{3^k} = \frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty \frac{1}{3^k(-2)^k} (-2x)^{r+k} = \frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty \left(-\frac{1}{6}\right)^k (-2x)^{r+k} = \frac13 \sum_{n=0}^\infty (-2x)^n \sum_{k=0}^n \left(-\frac{1}{6}\right)^k$$ where the sum over $k$ is again a geometric series.

One can show that both expressions for the coefficient of $x^n$ are equal, as they must be.

So the coefficient is $\frac{1}{21}(3^{-1}+(-1)^n(2)^{n+1}(3))$.
I still don't follow the substitution of n for r+k. How is that justified?

pasmith
Homework Helper
So the coefficient is $\frac{1}{21}(3^{-1}+(-1)^n(2)^{n+1}(3))$.
I still don't follow the substitution of n for r+k. How is that justified?
The series is absolutely convergent, so its terms can be re-ordered. Instead of taking them in the order (0,0), (0,1), (0,2), ..., (1,0), (1,1), .... one takes them in the order (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

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Ray Vickson
Homework Helper
Dearly Missed
So the coefficient is $\frac{1}{21}(3^{-1}+(-1)^n(2)^{n+1}(3))$.
I still don't follow the substitution of n for r+k. How is that justified?
Originally you have a sum over all pairs (r,k) with 0 ≤ r < ∞ and 0 ≤ k < ∞. This is just all the points in the lattice {(0,0), (1,0), (0,1), (1,1), (0,2) ,..... } in 2-dimensional (r,k)-space. Because of absolute convergence (at least, in a certain x-range) you can re-express the sum: sum over each of the ##45^o ## lines ##r+k = 0, 1, 2, \ldots##, and within each such line, sum over the lattice points contained in that portion of the line between the k and r axes. That is, for fixed ##r+k = n## you need to do the sum for (say) ##k = 0,1,2, \ldots, n##, with ##r = n - k## for each such ##k##. Why would you do that? The answer is: because you have terms ##x^{r+k}## in the sum, and you want to isolate the coefficient of ##x^{r+k}## for each fixed value ##r+k = n##.

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Thank you both very much for your help, very interesting and illuminating.