Formal power series coefficient problem

Question

Find the coefficient of:

$[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$

I have figured out that I need to use this identity:

$(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $

But I have no clue how to proceed with this, I have been stuck with this for hours please help.

Answer 1

$$ \begin{align} [x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37} &=[x^{32}](1+5x^6)^{-13}(1-8x^9)^{-37}\\ &+[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt] &=[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt] &=[x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37} \end{align} $$ The binomial theorem gives $$ \begin{align} (1+5x^2)^{-13} &=\sum_{j=0}^\infty\binom{-13}{j}\left(5x^2\right)^j\\ &=\sum_{j=0}^\infty\binom{j+12}{12}\left(-5x^2\right)^j\\ \end{align} $$ and $$ \begin{align} (1-8x^3)^{-37} &=\sum_{k=0}^\infty\binom{-37}{k}\left(-8x^3\right)^k\\ &=\sum_{k=0}^\infty\binom{k+36}{36}\left(8x^3\right)^k\\ \end{align} $$ Therefore, $$ \begin{align} [x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37} &=\sum_{2j+3k=10}\binom{j+12}{12}\binom{k+36}{36}(-5)^j8^k\\ &=\overbrace{\binom{17}{12}\binom{36}{36}(-5)^58^0}^{j=5,\,k=0}+\overbrace{\binom{14}{12}\binom{38}{36}(-5)^28^2}^{j=2,\,k=2}\\[15pt] &=83019300 \end{align} $$

---

Given

Coefficient[Series[(x+x^3)(1+5x^6)^-13(1-8x^9)^-37,{x,0,33}],x,33]

Mathematica returns 83019300

Answer 2

You know that

$$(1+5x^6)^{-13}=\sum_{n\ge 0}\binom{n+12}{12}\left(-5x^6\right)^n=\sum_{n\ge 0}\binom{n+12}{12}(-1)^n5^nx^{6n}\tag{1}$$

and

$$(1-8x^9)^{-37}=\sum_{n\ge 0}\binom{n+36}{36}\left(8x^9\right)^n=\sum_{n\ge 0}\binom{n+36}{36}8^nx^{9n}\;.\tag{2}$$

The coefficient of $x^{33}$ in $(x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$ is the sum of the coefficients of $x^{33}$ in $x(1+5x^6)^{-13}(1-8x^9)^{-37}$ and $x^3(1+5x^6)^{-13}(1-8x^9)^{-37}$. Clearly

$$[x^{33}]x(1+5x^6)^{-13}(1-8x^9)^{-37}=[x^{32}](1+5x^6)^{-13}(1-8x^9)^{-37}$$

and

$$[x^{33}]x^3(1+5x^6)^{-13}(1-8x^9)^{-37}=[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\;,$$

so you need to find the coefficients of $x^{30}$ and $x^{32}$ in $(1+5x^6)^{-13}(1-8x^9)^{-37}$. If for convenience we write

$$(1+5x^6)^{-13}=\sum_{n\ge 0}a_nx^n\qquad\text{and}\qquad(1-8x^9)^{-37}=\sum_{n\ge 0}b_nx^n\;,$$

then we know that the coefficients of $30$ and $32$ in $(1+5x^6)^{-13}(1-8x^9)^{-37}$ are

$$\sum_{k=0}^{30}a_kb_{30-k}\qquad\text{and}\qquad\sum_{k=0}^{32}a_kb_{32-k}\;,\tag{3}$$

respectively. Now use $(1)$ and $(2)$ to evaluate these coefficients. Note that the only powers of $x$ with non-zero coefficients in $(1)$ are those whose exponents are multiples of $6$, and the only powers of $x$ with non-zero coefficients in $(2)$ are those whose exponents are multiples of $9$, so most of the terms in $(3)$ will be $0$. (In fact, with a little thought you can determine that one of the coefficients is $0$ without actually doing any arithmetic.)