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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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  • $\begingroup$ Perfect, thank you! I came up with C(11,6) as my answer $\endgroup$ – audiFanatic Nov 6 '12 at 4:49
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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 $$a_6=\binom{6+6-1}{6}=\binom{11}{6}$$

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