Coefficient of Generating Function Series

Question

Let

$S(x) = \frac{x^2 + x}{(1-x)^3}$

What is the coefficient of $x^2$ in the generating function series for $S(x)$ ?

Answer 1

If $S(x)=a_0+a_1x+a_2x^2+.....$, then $a_2=\frac{S''(0)}{2}$

Answer 2

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

$$x+4x^2+\ldots$$

Answer 3

If you know that

$$\frac1{(1-x)^{m+1}}=\sum_{n\ge 0}\binom{n+m}nx^n\;,$$

you can take $m=2$ to get

$$\frac1{(1-x)^3}=\sum_{n\ge 0}\binom{n+2}2x^n=1+3x+6x^2+\ldots\;.$$

Then you have

$$(x+x^2)(1+3x+6x^2+\ldots)\;,$$

and it’s clear that the only $x^2$ terms are $x\cdot3x$ and $x^2\cdot1$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \bracks{x^{2}}\bracks{x^{2} + x \over \pars{1 - x}^{3}} & = \bracks{x^{2}}\bracks{{x^{2} \over \pars{1 - x}^{3}} + {x \over \pars{1 - x}^{3}}} \\[5mm] & = \underbrace{\bracks{x^{0}}\bracks{1 \over \pars{1 - x}^{3}}}_{\ds{1}}\ +\ \underbrace{\bracks{x^{1}}\bracks{1 \over \pars{1 - x}^{3}}}_{\ds{3}}\ =\ \bbx{\ds{4}} \end{align}

because $\ds{{1 \over \pars{1 - x}^{3}} = \color{#f00}{1} + \color{#f00}{3}x + \mrm{O}\pars{x^{2}}}$