Compute the power series of $\frac{2-6x+x^2}{1-3x}$

Question

I'm struggling with the following:

Suppose that we have a formal power series $\sum_{n \geq 0} a_nx^n$ which is equal to the fraction $\frac{2-6x+x^2}{1-3x}.$ I want to find an explicit formula for $a_n.$

Here is what I have done so far: I realise that $(1-x)^{-n} = \sum_{k \geq 0} {n + k - 1 \choose k}x^k,$ so I can say that $\frac{2-6x+x^2}{1-3x} = (2-6x+x^2)(1-3x)^{-1} = (2-6x+x^2)\sum_{k \geq 0} {k \choose k} x^k.$

From here we obtain

$ \sum_{k \geq 0} 2x^{k} - \sum_{k \geq 0} 6x^{k+1} + \sum_{k \geq 0} x^{k+2}.$

Here I have no clue how to continue or if I'm even on the right track. Hints and or solutions are very welcome

Answer 1

Your method is perfect, except that you forgot the factor $3$: $(1-3x)^{-1}=\sum_{k=0}^\infty({\color{red}3}x)^k$, hence

$$\frac{2-6x+x^2}{1-3x}=\sum_{k\ge0}2\cdot3^kx^k-\sum_{k\ge0}6\cdot3^kx^{k+1}+\sum_{k\ge0}3^kx^{k+2}$$ $$=\sum_{k\ge0}2\cdot3^kx^k-\sum_{m\ge1}6\cdot3^{m-1}x^m+\sum_{n\ge2}3^{n-2}x^n$$ $$=2+\sum_{n\ge2}3^{n-2}x^n.$$

Remark. If you first notice that the numerator is $2(1-3x)+x^2$, you get directly:

$$\frac{2-6x+x^2}{1-3x}=2+\sum_{k\ge0}3^kx^{k+2}=2+\sum_{n\ge2}3^{n-2}x^n$$

Answer 2

I will provide another method by directly computing the $n$-th derivative at $x=0$ as it is a power series centered at $x=0$. By partial fracition, we have $$\dfrac{2-6x+x^2}{1-3x}\equiv\dfrac{17-3x}{9}-\dfrac{1}{9(3x-1)}$$ Hence we have \begin{align*}f^{(0)}(0)&=2\\ f^{(1)}(0)&=\dfrac{-3}{9}+\dfrac{3}{9(-1)^2}\\ f^{(n)}{(0)}&=\left.\dfrac{d^{n}}{dx^n}\dfrac{-1}{9(3x-1)}\right|_{x=0}=\dfrac{3^{n-2}(-1)^{n+1}(n!)}{(0-1)^{n+1}}=n!3^{n-2},n\ge2\end{align*} Hence, we can set $a_n=\dfrac{f^{(n)}(0)}{n!}$ to obtain our desired power series.