Determining the coefficient of $x^6$ in a generating function

Question

I've been trying to find the coefficient of $x^6$ in the generating function:

$$f(x) = \dfrac 1{x(2x-1)^2}$$

I tried to separate the $\dfrac 1x$ and get $\dfrac 1x\cdot \dfrac{1}{(2x-1)^2}$ and then go from there, but I don't know what to do after that.

Do I factor the denominator of $\dfrac 1{(2x-1)^2}$ by $-2$? What can I do with $\dfrac 1x$? I've never seen $\dfrac 1x$ before in a generating function so I don't know what to do with it.

The answer should not contain a binomial term. I think I should use the 1/(1-x)^n form but I don't know how to do i with 1/x. I know with x I subtract one from [x^6], does that mean I should add 1 to [x^6]?

UPDATE

So I added 1 to [x^6] and now I'm trying to find out how to rewrite 1/(2x-1)^2. If I factor 2 out I'd get -1/4 * 1/(1/2 - x)^2. Is that correct? But I don't know what to do here since it's not in the form 1/(1-x). What should I do with the 1/2?

Answer 1

We obtain \begin{align*} \color{blue}{[x^6]}\color{blue}{\frac{1}{x(2x-1)^2}}&=[x^7]\frac{1}{(1-2x)^2}\tag{1}\\ &=[x^7]\sum_{j=0}^\infty\binom{-2}{j}(-2x)^j\tag{2}\\ &=[x^7]\sum_{j=0}^\infty(j+1)(2x)^j\tag{3}\\ &\,\,\color{blue}{=2^{10}}\tag{4} \end{align*}

Comment:

• In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (2) we apply the binomial series expansion.

• In (3) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and $\binom{p}{p-1}=p$.

• In (4) we select the coefficient of $x^7$.

Answer 2

$$\frac{1}{1-2x} = \sum_{k \ge 0} (2x)^k$$

$$\frac{1}{(1-2x)^2} = \sum_{k \ge 0} (k+1) (2x)^k$$

$$\frac{1}{x(1-2x)^2} = \sum_{k \ge -1} (k+2) 2^{k+1} x^k$$