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Find the $x^{24}$ coefficient in the expansion of $\left(1-x^6\right)^{-2}\left(1-x^3\right)^{-1}(1-x)^{-1}$
Option: 1
$225$

Option: 2
$55$

Option: 3
$210$

Option: 4
$125$

Answers (1)

The given rational number can be written as

$\left(1+x^6+x^{12}+\ldots\right)\left(1+x^6+x^{12}+\ldots\right)\left(1+x^3+x^6+\ldots\right)\left(1+x^2+x^3+\ldots\right) .$

The coefficient $c_2_4$ we are after therefore is the number of nonnegative integer solutions to

$6 k_1+6 k_2+3 k_3+k_4=24$

$\text { or } 6 k_1+6 k_2+3 k_3 \leq 24 \text {, which is the same as }$

$2\left(k_1+k_2\right)+k_3 \leq 8 .$

If $k_1+k_2=: j \geq 0$ each given value of $j$ can be realized by $k_1$ and $k_2$ in $j+1$ ways, and then $k_3$ may take $8-2 j+1$ different values. It follows that

$c_{24}=\sum_{j=0}^4(j+1) \cdot(9-2 j)=1 \cdot 9+2 \cdot 7+3 \cdot 5+4 \cdot 3+5 \cdot 1=55 .$

Posted by

Ramraj Saini

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