Find the sum of the coefficients of $x^{20}$ and $x^{21}$ in the power series expansion of $\frac 1{(1-x^3)^4}$.

Question

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!

Answer 1

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 $.

Answer 2

The power series expansion of $\frac 1{1-y}$ is $1+y+y^2+y^3+\cdots$, which we can think of as the generating function for all nonnegative integers. Substituting $x^3$ for $y$, we get $$\frac 1{1-x^3} = 1 + x^3 + x^6 + x^9 + \cdots$$ and thus $$\frac 1{(1-x^3)^4} = (1 + x^3 + x^6 + x^9 + \cdots)^4.$$ The coefficient of $x^k$ counts the number of ways to write $k$ as the sum of an ordered quadruple of nonnegative multiples of $3$. If $k$ is not itself a multiple of $3$, then there are no such quadruples; thus the coefficient of $x^{20}$ is $0$.

For the coefficient of $x^{21}$, we note that solving $$a+b+c+d=21$$ with $a,b,c,d$ all multiples of $3$ is equivalent via a substitution to solving $$a'+b'+c'+d'=7$$ in nonnegative integers. From our work on distribution problems, we know that there are $\binom{7+4-1}{4-1}=\binom{10}3=120$ ways to do this.

Therefore, the sum of the coefficients of $x^{20}$ and $x^{21}$ in the expansion of $\frac 1{(1-x^3)^4}$ is $0+120=\boxed{120}$.

Solution from AoPS