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

Mathematics Asked on December 23, 2021

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}} = sumlimits_{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.

3 Answers

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$

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


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}

Answered by Felix Marin on December 23, 2021

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 $n_iinmathbb{W}$.

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

Answered by SarGe on December 23, 2021

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 $$sum_{c=0}^8 12 - leftlceil frac{3}{2} c rightrceil + 1 = 13(9) - (0 + 2 + 3 + 5 + 6 + 8 + 9 + 11 + 12) = 61$$ such solutions, and this is the desired coefficient.

Answered by heropup on December 23, 2021

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