Here p can be $0, 1, 2...$ and $n$ can be $0,1,2,...$ I just want to get the generalized relation for finding the coefficient of $x^2$ or $x^3$ and so on in expansion of multinomials like $(1+x+x^2)^5$ and so on. Thanks!
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$\begingroup$ In your polynomial $(1+x+x^2)^5$, if we expand into five identical brackets, there are two brackets that we must choose $x^1$ from, and the rest are $x^0$s. Additionally, there are 5 ways to choose an $x^2$ in the 5 brackets. Therefore, there are ${5 \choose 2} + 5$ ways to choose $x^2$, and that is the coefficient of $x^2$. $\endgroup$– Toby MakCommented Jun 22, 2017 at 8:46
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$\begingroup$ This is $\sum_{\{j,k\in\mathbb{N}^2 / 2j+k=p\}}a^jb^kc^{n-j-k}$. Not sure you can get a closed formula. $\endgroup$– EvargaloCommented Jun 22, 2017 at 8:46
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$\begingroup$ @TobyMak : did you forget five terms made of one $x^2$ and four $1$'s ? The answer should be $\binom{5}{2}+\binom{5}{1}=15$. $\endgroup$– EvargaloCommented Jun 22, 2017 at 8:49
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1$\begingroup$ Corrected the post. $\endgroup$– Toby MakCommented Jun 22, 2017 at 8:50
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$\begingroup$ Correcting my comment above:This is $\sum_{\{j,k\in\mathbb{N}^2 / 2j+k=p\}}\binom{n}{j}\binom{n-j}{k}a^jb^kc^{n-j-k}$. Not sure you can get a closed formula. $\endgroup$– EvargaloCommented Jun 22, 2017 at 8:53
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1 Answer
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Choosing $i$ copies of $ax^2$, $j$ copies of $bx$ with $i+j\le n$ and $n-i-j$ copies of $c$ gives $$a^ib^jc^{n-i-j}x^{2i+j},$$an $x^p$ term iff $2i+j=p$. For such a choice of $i,\,j$ there are $$\binom{n}{i}\binom{n-i}{j}=\frac{n!}{i!j!\left( n-i-j\right)!}$$ways to select these factors. The $x^p$ coefficient is therefore $$\sum_{0\le i,\,j\le n,\,2i+j=p}\frac{n!}{i!j!\left( n-i-j\right)!}a^ib^jc^{n-i-j}.$$In general, this cannot be simplified further.
