Find the coefficient of $x^{24}$ in the binomial equation

Question

Find the coefficient of $x^{24}$ in the equation ${\left( {1 - x} \right)^{ - 1}}.{\left( {1 - {x^2}} \right)^{ - 1}}.{\left( {1 - {x^3}} \right)^{ - 1}}$

My approach is as follow

The equation used is ${\left( {1 - x} \right)^{ - n}} = \sum\limits_{r = 0}^\infty {{}^{n + r - 1}{C_r}{x^r}} $

Then after expanding I get the following ${\left( {1 - x} \right)^{ - 3}}.{\left( {1 + x} \right)^{ - 1}}.{\left( {1 + x + {x^2}} \right)^{ - 1}}$

As ${\left( {1 - x} \right)^{ - 3}}$ is defined by the above formula I can expand it, but how I will do the expansion for other terms. Any other shortcut method.

Answer 1

If you think about the product as $$\frac{1}{(1-x)(1-x^2)(1-x^3)} = (1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots),$$ it becomes clear that the coefficient of $x^{24}$ is given by the number of ways to choose nonnegative integers $a,b,c$ such that $a + 2b + 3c = 24.$ Fortunately, these are not too difficult to enumerate since $24$ is small. The best way to proceed is to first choose $c$; i.e., for each $c \in \{0, 1, 2, \ldots, 8\}$, solve $$a + 2b = 24 - 3c = 3(8-c).$$ This in turn requires $$a = 3(8-c) - 2b \ge 0,$$ so $b \in \{0, 1, \ldots, 12 - \lceil \tfrac{3}{2}c \rceil\}$. So that means there are $$\sum_{c=0}^8 12 - \left\lceil \frac{3}{2} c \right\rceil + 1 = 13(9) - (0 + 2 + 3 + 5 + 6 + 8 + 9 + 11 + 12) = 61$$ such solutions, and this is the desired coefficient.

Answer 2

I suggest don't expand the terms. There are three expressions in multiplication which give power of $x$ as multiple of $1$, $2$ and $3$ respectively. Let the exponent of $x$ from each expression be $n_1, \, n_2$ and $n_3$ respectively where $n_i\in\mathbb{W}$.

So, you've to find no. of non-negative integral solutions of $n_1+2n_2+3n_3=24$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\bracks{\cdots}}$ "inside the sums" is an Iverson Bracket.

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\begin{align} &\bbox[10px,#ffd]{\bracks{x^{24}}\pars{1 - x}^{-1}\pars{1 - x^{2}}^{-1}\pars{1 - x^{3}}^{-1}} = \bracks{x^{24}}\sum_{i = 0}^{\infty}x^{i}\sum_{j = 0}^{\infty}x^{2j} \sum_{k = 0}^{\infty}x^{3k} \\[5mm] = & \sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty} \bracks{i + 2j + 3k = 24} = \sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty} \bracks{24 - 2j - 3k \geq 0} \\[5mm] = &\ \sum_{j = 0}^{\infty}\sum_{k = 0}^{\infty} \bracks{j \leq 12 - {3 \over 2}\,k} = \sum_{k = 0}^{\infty}\sum_{j = 0}^{\infty} \bracks{j \leq 12 - {3 \over 2}\,k} \bracks{12 - {3 \over 2}\,k \geq 0} \\[5mm] = &\ \sum_{k = 0}^{8}\sum_{j = 0}^{\infty} \bracks{j \leq 12 - {3 \over 2}\,k} \\[1cm] = &\ \underbrace{\sum_{j = 0}^{12}1}_{\ds{k\ =\ 0}}\ +\ \underbrace{\sum_{j = 0}^{10}1}_{\ds{k\ =\ 1}}\ + \underbrace{\sum_{j = 0}^{9}1}_{\ds{k\ =\ 2}}\ +\ \underbrace{\sum_{j = 1}^{7}1}_{\ds{k\ =\ 3}}\ +\ \underbrace{\sum_{j = 0}^{6}1}_{\ds{k\ =\ 4}}\ +\ \underbrace{\sum_{j = 0}^{4}1}_{\ds{k\ =\ 5}}\ +\ \underbrace{\sum_{j = 0}^{3}1}_{\ds{k\ =\ 6}} \\[2mm] & + \underbrace{\sum_{j = 0}^{1}1}_{\ds{k\ =\ 7}}\ +\ \underbrace{\sum_{j = 0}^{0}1}_{\ds{k\ =\ 8}} \\[1cm] = &\ 13 + 11 + 10 + 8 + 7 + 5 + 4 + 2 + 1 = \bbx{61} \end{align}