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Find the constant term in the expansion of $$\left(x^2 + x + \frac{1}{x} + \frac{1}{x^2}\right)^{15}$$

I know the answer is

$$ \sum _{k = 0}^{5} \binom{15}{5+k}\binom{15}{3k} =68974906 $$

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    Just apply the multinomial theorem. Or are you curious as to how to find the sum? – 2017-02-27
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    I tried that, it gives a 4 variable sum with constraint. From there I can't proceed. – 2017-02-27
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    @dark32 Please explain. How are you getting that $4$-variable sum? Note that you only have to consider the products in which the exponents sum up to $0$. – 2017-02-27

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$$ \begin{align} \left[x^0\right]\left(x^2+x+\frac1x+\frac1{x^2}\right)^{15} &=\left[x^0\right]\left(\frac1{x^{30}}(x+1)^{15}\left(x^3+1\right)^{15}\right)\\ &=\left[x^{30}\right]\left((x+1)^{15}\left(x^3+1\right)^{15}\right)\\ &=\left[x^{30}\right]\sum_{k,j}\binom{15}{j}\binom{15}{k}x^{3k+j}\\ &=\sum_{3k+j=30}\binom{15}{j}\binom{15}{k}\\ &=\sum_{k=5}^{10}\binom{15}{30-3k}\binom{15}{k}\\ &=\sum_{k=0}^5\binom{15}{15-3k}\binom{15}{k+5}\\ &=\sum_{k=0}^5\binom{15}{3k}\binom{15}{k+5}\\ \end{align} $$