Multinomial Coefficient of $x^{1397}$ in expansion of $(x^3+x^4+x^5+...)^6$

Question

I have the following problem:

Find the Coefficient of $x^{1397}$ in expansion of $(x^3+x^4+x^5+...)^6$

I know how to solve these kind of questions using Multinomial Theorem but since the polynomial in this one is infinite I’m lost!

Thanks in advance.

Answer 1

This is the number of solutions to the equation $Z_1 + \dots + Z_6 = 1397$ where $Z_1$, $\dots$, $Z_6$ are positive integers $\geq 3$.

Subtracting three from each of the $Z_i$'s, each solution corresponds to a solution of $$X_1 + \dots + X_6 = 1397 - (6 \cdot 3) = 1379$$ where each $X_i$ is a non-negative integer.

In general, the number of solutions to $X_1 + \dots + X_k = n$ for integer $X_i \geq 0$ is $\binom{n + k - 1}{k-1}$. So in your particular case, $k = 6$ and $n = 1379$; the answer is $\binom{1384}{5} = 42010498234776$.

Answer 2

You are finding the coefficient of $x^{1397 - 3 \times 6} = x^{1379}$ of $(1 + x + x^2 + \cdots)^{6}$. Since you don't have to care about the term after $x^{1379}$ in $1+ x+ x^2+ \cdots$, you are finiding the coefficient of $x^{1379}$ of $(1+ x + x^2 + \cdots + x^{1379})^6$.

If you are not bounded to use the multinomial theorem, my suggestion is to find the taylor series of $(1+x+\cdots)^6 = \frac{1}{(1-x)^6}$ and find the $1379$th coefficient.