How to do a multinomial theorem sum faster

Question

For example we have this question : Find the coefficient of $x^6$ in the following

$\frac{\left(x^{2}+x+2\right)^{9}}{20}$

So using multinomial Theorem which is this :

$\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{n}=\sum_{b_{1}+b_{2}+\cdots+b_{k}=n}\left(\begin{array}{c} n \\ b_{1}, b_{2}, b_{3}, \dots, b_{k} \end{array}\right) \prod_{j=1}^{k} x_{j}^{b_{j}}$

I have to first find out the combinations of power for which I am going to get $x^6$ then i have to calculate the multinomial coefficient multiplied by the $2$ raised to some power for all of those combinations.

Is there an easier way to somehow change this question into some other question then use Binomial or some other theorem to solve it quickly ?

Answer 1

Extracting a сomplete square can save some time. For example, from $x^2+x+2=\frac{(2x+1)^2+7}{4}$, we get $$[x^6]\, (x^2+x+2)^9 = \frac{1}{4^9} \sum_{k=0}^9 \binom{9}{k} \binom{2k}{6} 2^6 7^{9-k}.$$

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Alternatively, we can employ the factorization $x^2+x+2 = (x-\alpha_1)(x-\alpha_2)$, where $\alpha_{1,2}=\frac{-1\pm I\sqrt{7}}2$, to get $$[x^6]\, (x^2+x+2)^9 = \sum_{i=0}^6 \binom9i\binom9{6-i}(-\alpha_1)^{9-i} (-\alpha_2)^{3+i}.$$

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Yet another approach is to represent $x^2+x+2=(x^2+x)+2$ and notice that $[x^6]\,(x^2+x)^k = \binom{k}{6-k}$. Then $$[x^6]\, (x^2+x+2)^9 = \sum_{k=0}^9 \binom9k \binom{k}{6-k}2^{9-k}.$$