Find sequence for given generating function $\frac{ \frac{3x}{2} + \frac{3}{2}}{ 3 - x }$

Question

I have generating function

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

and I need to find a sequence from it.

This was my approach:

$$ \frac{ \frac{3x}{2} + \frac{3}{2}}{ 3 - x } = \frac{-3}{2} + 2 \cdot \frac{1}{1 - \frac{x}{3}} $$

$ \frac{1}{1 - \frac{x}{3}} $ is generating function for $a_n = \frac{1}{3^n}$ so I though that the whole sequence $(c_n)_0^\infty$ would be:

$$c_n = \frac{-3}{2} + \frac{2}{3^n} $$

but according to wolfram it is wrong, where I am doing a mistake?

Answer 1

$$\begin{align*} \frac32\cdot\frac{x+1}{3-x}&=\frac12\cdot\frac{x+1}{1-\frac{x}3}\\ &=\frac12\left(x\sum_{n\ge 0}\left(\frac13\right)^nx^n+\sum_{n\ge 0}\left(\frac13\right)^nx^n\right)\\ &=\frac12\left(\sum_{n\ge 0}\left(\frac13\right)^nx^{n+1}+\sum_{n\ge 0}\left(\frac13\right)^nx^n\right)\\ &=\frac12\left(\sum_{n\ge 1}\left(\frac13\right)^{n-1}x^n+\sum_{n\ge 1}\left(\frac13\right)^nx^n+1\right)\\ &=\frac12\left(\sum_{n\ge 1}\left(\left(\frac13\right)^{n-1}+\left(\frac13\right)^n\right)x^n+1\right)\\ &=\frac12+\frac12\sum_{n\ge 1}\left(\frac43\right)\left(\frac13\right)^{n-1}x^n\\ &=\frac12+2\sum_{n\ge 1}\left(\frac13\right)^nx^n\;, \end{align*}$$

so $a_0=\dfrac12$, and $a_n=\dfrac2{3^n}$ for $n\ge 1$.

Answer 2

Write $$\frac{3}{6}\frac{x}{1 - \frac{x}{3}} + \frac{3}{6} \frac{1}{1 - \frac{x}{3}}$$

and use that $$\frac{1}{1 - \frac{x}{3}} = \sum_{n=0}^{\infty} \bigg( \frac{x}{3}\bigg)^n$$