Finding the coefficent of Binomial series

Question

I am not sure how to get the coefficient of the following,

$$[x^n]\frac{(1+x)^n}{1-x}$$

The answer appears to be $2^n$ but I am not sure how to get it. Please help! :)

Answer 1

HINT:

$$\begin{align*} \frac{(1+x)^n}{1-x}&=\frac1{1-x}\cdot(1+x)^n\\ &=\left(\sum_{k\ge 0}x^k\right)\sum_{k\ge 0}\binom{n}kx^k \end{align*}$$

(It’s all right to take the second sum over $k\ge 0$, since $\binom{n}k=0$ when $k>n$.) Now take the Cauchy product of the two series and see what the coefficient of $x^n$ is. This result is a special case of a useful fact that I’ve spoiler-protected below.

A useful fact: if $g(x)=\sum_{k\ge 0}a_kx^k$ is the generating function for the sequence $\langle a_n:n\in\Bbb N\rangle$, then $\frac{g(x)}{1-x}$ is the generating function for the convolution of the sequence $\langle 1,1,1,\ldots\rangle$ with the sequence $\langle a_n:n\in\Bbb N\rangle$, which is the sequence $\langle a_0,a_0+a_1,a_0+a_1+a_2,\ldots\rangle$ of partial sums of the sequence $\langle a_n:n\in\Bbb N\rangle$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \bracks{x^{n}}{\pars{1 + x}^{n} \over 1 - x} & = 2^{n}\bracks{x^{n}}{\bracks{1 - \pars{1 - x}/2}^{\,n} \over 1 - x} \\[5mm] & = 2^{n}\braces{\overbrace{\bracks{x^{n}}\pars{1 - x}^{-1}}^{\ds{=\ 1}}\ +\ \sum_{k = 1}^{n}{n \choose k}{\pars{-1}^{k} \over 2^{k}}\ \overbrace{\bracks{x^{n}}\pars{1 - x}^{k - 1}}^{\ds{=\ 0}}} = \bbx{\ds{2^{n}}} \end{align}