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

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!

• In the $...$, do you think powers of $x$ beyond $1397$ matter? Convert the problem into a combinatorial problem by looking at six numbers summing to $1397$. May 29 '20 at 6:14

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 $$\binom{1384}{5} = 42010498234776$$.
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.