I know we can factor the bottom using difference of cubes to get $\frac{1}{(1-x)^4(1+2x^3+x^6)^4}$. Then we can take out the $(1-x)^4$ to get $\frac{1}{(1+2x^3+x^6)^4} \cdot [\binom 33 + \binom {4}{3}x + \binom 53x^2 \cdots]$. I don't know how to simplify any further - please tell me if I did something wrong. Thanks!
Find the sum of the coefficients of $x^{20}$ and $x^{21}$ in the power series expansion of $\frac 1{(1-x^3)^4}$.
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power-series
binomial-coefficients
generating-functions
1 Answers
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Since $(1-x)^{-n} =\sum_{k=0}^{\infty} \binom{n+k-1}{n-1}x^k $ (see https://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalized_binomial_theorem), $(1-x^m)^{-n} =\sum_{k=0}^{\infty} \binom{n+k-1}{n-1}x^{km} $.
Therefore $(1-x^3)^{-4} =\sum_{k=0}^{\infty} \binom{k+3}{3}x^{3k} $.
Since there is no term with $x^{20}$ and the term with $x^{21}$ corresponds to $k = 7$, the result is $\binom{10}{3} =120 $.