Find the coefficient of $x^k$

Question

Find the coefficient of $x^k$ in $$\frac{1}{(1+x)(2-9x)}.$$ This problem is in chapter of Algebraic Tools, using generating functions.

Answer 1

Here is a method to get the answer without assuming the development of $\frac 1{ax+b}$ in advance.

We have $\quad f(x)=\dfrac 1{(1+x)(2-9x)}\iff (-9x^2-7x+2)f(x)=1$

Now let's set $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$ and try to find a relation between the coefficients.

$\sum\limits_{n=0}^{\infty}(-9a_nx^{n+2}-7a_nx^{n+1}+2a_nx^{n})=1$

$\iff-\sum\limits_{n=2}^{\infty}9a_{n-2}x^n-\sum\limits_{n=1}^{\infty}7a_{n-1}x^n+\sum\limits_{n=0}^{\infty}2a_nx^n=1$

$\iff\bigg(-7a_0x+2a_0+2a_1x\bigg)+\sum\limits_{n=2}^{\infty}\bigg(-9a_{n-2}-7a_{n-1}+2a_n\bigg)x^n=1$

Identifying the coefficients to the constant series $1$ gives:

$\begin{cases} 2a_0=1\\ -7a_0+2a_1=0\\ -9a_{n-2}-7a_{n-1}+2a_n=0\quad\forall n\ge 2\\ \end{cases}$

The characteristic equation of this linear induction relation

$2r^2-7r-9=(r+1)(2r-9)=0\quad$ has roots $-1$ and $\frac 92$.

Thus $a_n=A(-1)^n+B(\frac 92)^n$

Solving for initial terms $a_0=\frac 12,\ a_1=\frac 74$ gives $A=\frac 1{11}$ and $B=\frac 9{22}$

Finally $$a_n=\dfrac{(-1)^n+\left(\frac 92\right)^{n+1}}{11}$$

Answer 2

Using partial fractions, we have $$\frac{1}{(1+x)(2-9x)} = \frac{1}{11} \cdot \frac{1}{x+1} - \frac{9}{11} \cdot \frac{1}{9x-2}.$$ From here, use infinite geometric series to expand the fractions, and add the terms.

Answer 3

If you prefer to avoid computing derivatives then you could rewrite it as $$\frac{1}{(1+x)(2-9x)}=\frac{1}{11} \left (\frac{1}{1+x}+\frac{9}{2-9x} \right)$$ Then $$\frac{1}{1+x}= \sum_{k\ge 0} (-x)^k$$ $$\frac{9}{2-9x}=\frac{9}{2} \frac{1}{1-\frac{9}{2}x} = \sum_{k\ge 0} \frac{9^{k+1}x^k}{2^{k+1}}$$

Thus you get that the coefficient of $x^k$ is

$$\frac{ 9^{k+1}- (-2)^{k+1}}{11\cdot 2^{k+1}}$$