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Find coefficient of $x^{18}$ in $(1+x^3+x^6+x^9+\cdots)^6 $

Normally, to to this I would look for a pattern that follows the following basic expansions in this powerpoint: http://academics.smcvt.edu/jellis-monaghan/combo2/Archive/Combo%20s03/class%20notes%20s03/seection6_2.ppt

However, what's throwing me off is the doubling exponents. I'm not quite sure how exactly to deal with them. Any help would be appreciated.

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Try substituting $x^3$ with, say $y$. Then can you do it?

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    Perfect, thank you! I came up with C(11,6) as my answer2012-11-06
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$(1+x^3+x^6+x^9+\cdots)^6 =\left(\frac{1}{1-x^3}\right)^6=(1-x^3)^{-6}=$ $=\sum_{k=0}^{\infty}\binom{6+k-1}{k}x^{3k}=\sum_{k=0}^{\infty}a_kx^{3k}\Rightarrow a_k=\binom{6+k-1}{k}$

$x^{18}=x^{3k}\Rightarrow 3k=18\Rightarrow k=6$ and coefficient is $a_6=\binom{6+6-1}{6}=\binom{11}{6}$