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I've been asked to compute

$[x^n]\frac{1}{(1-2x)(1+3x^2)}$

where $[x^n]$ is the coefficient of the series.

I've recognized that it is a geometric series, and have been able to put it into this form:

$[x^n]\sum_{n\ge0} (2x)^n \sum_{n\ge0} (-3x^2)^n$

However, by reducing and simplifying I get to this point:

$[x^n]\sum_{n\ge0} (-6)^n x^{3n}$

From here I'm lost as to where I can go.

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I think you may be getting confused about the $n$ variable. You used it for two different things in the same expression. It may help to give different names to the index of your two summations in your second expression. For example, $\sum_{n \ge 0}$ for the first and $\sum_{m \ge 0}$ for the second. –  Snowball Sep 25 '12 at 21:50
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3 Answers

up vote 3 down vote accepted

Expanding as you did is a good first step. We end up having $$\sum_{n=0}^\infty (2x)^n\sum_{m=0}^\infty (-3x^2)^m = \sum_{n,m=0}^\infty 2^n(-3)^m x^{2m+n}$$ where the latter sum happens over all $m\ge 0,\ n\ge 0$. If we rewrite the sum in terms of $k=2m+n$ then $$\sum_{k=0}^\infty \left(\sum_{2m+n=k} 2^n(-3)^m\right)x^k$$ therefore the coefficient for $[x^k]$ you are after is $$\sum_{2m+n=k}2^n(-3)^m$$ which sums over all pairs of $m, n\ge 0$ such that $2m+n=k$. We rewrite the sum as $$\sum^{\lfloor\frac{k}{2}\rfloor}_{m=0}2^{k-2m}(-3)^{m}=2^k\sum^{\lfloor\frac{k}{2}\rfloor}_{m=0}\left(\frac{-3}{4}\right)^m$$ The latter is a geometric series which easily evaluates to $$\frac{2^{k+2}}{7}\left[1 - \left(-\frac{3}{4}\right)^{\lfloor\frac{k+2}{2}\rfloor}\right]$$

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I'm a bit confused as to why the top of the sum towards the end is k/2 ? –  Overload119 Sep 25 '12 at 22:30
    
Consider the pairs of numbers for which $2m + n =k$. Let's use the concrete example $k=7$. Then we have $(m=0, k =7)$, $(m=1, k=5)$, ... continuing onwards you'll see that $m$ peaks at $\lfloor\frac{k}{2}\rfloor$. To see it another way, the highest $m$ can be is if $n=0 \implies 2m=k \implies m \le \frac{k}{2}$. –  EuYu Sep 25 '12 at 22:33
    
Makes sense thanks! –  Overload119 Sep 25 '12 at 22:41
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You might want to aim at partial fraction decomposition

$$\frac{1}{{\left( {1 - 2x} \right)\left( {1 + 3{x^2}} \right)}} = \frac{A}{{1 - 2x}} + \frac{{Bx + C}}{{1 + 3{x^2}}}$$

and then find the coefficients separately.

After equating coefficients, you should get the system $$\eqalign{ & A + C = 1 \cr & B - 2C = 0 \cr & 3A - 2B = 0 \cr} $$ from where $$\eqalign{ & A = 4/7 \cr & B = 6/7 \cr & C = 3/7 \cr} $$

$$\frac{1}{{\left( {1 - 2x} \right)\left( {1 + 3{x^2}} \right)}} = \frac{1}{7}\left( {\frac{4}{{1 - 2x}} + \frac{{6x + 3}}{{1 + 3{x^2}}}} \right)$$

now use the usual expansions

$$\eqalign{ & \frac{4}{{1 - 2x}} = 4\sum\limits_{n = 0}^\infty {{2^n}{x^n}} \cr & \frac{1}{{1 + 3{x^2}}} = \sum\limits_{n = 0}^\infty (-1)^n {{{ { 3} }^{n}}{x^{2n}}} \cr} $$

You might want to consider odd and even terms separately.

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Find the partial fraction decomposition of the function and work with it:

$$\begin{align*} \frac{1}{(1-2x)(1+3x^2)}&=\frac47\cdot\frac1{1-2x}+\frac37\cdot\frac{1+2x}{1+3x^2}\\ &=\frac47\sum_{k\ge 0}(-1)^k2^kx^k+\frac37(1+2x)\sum_{k\ge 0}3^kx^{2k}\\ &=\frac47\sum_{k\ge 0}(-1)^k2^kx^k+\frac37\sum_{k\ge 0}3^kx^{2k}+\frac67\sum_{k\ge 0}3^kx^{2k+1}\;.\tag{1} \end{align*}$$

The coefficient of $x^n$ in $(1)$ is

$$\begin{align*} &\begin{cases} \frac47\cdot2^n+\frac373^{n/2},&\text{if }n\text{ is even}\\\\ -\frac47\cdot2^n+\frac673^{(n-1)/2},&\text{if }n\text{ is odd} \end{cases}\\\\ &\qquad=\begin{cases} \frac17\left(2^{n+2}+3^{(n+2)/2}\right),&\text{if }n\text{ is even}\\\\ \frac17\left(-2^{n+2}+2\cdot3^{(n+1)/2}\right),&\text{if }n\text{ is odd} \end{cases}\\\\ &\qquad=\frac17\left((-1)^n2^n+\left(1-(-1)^n\right)\cdot3^{\lfloor n/2\rfloor+1}\right)\\\\ &\qquad=\frac17\left((-1)^n\left(2^n-3^{\lfloor n/2\rfloor-1}\right)+3^{\lfloor n/2\rfloor-1}\right)\;, \end{align*}$$

where I’ve given you a choice of reasonable ways to write the expression.

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