The teacher briefly glossed over the multinomial theorem and then dropped this seemingly monstrous homework problem on us:
Find the coefficient of $x^{12}$ in the expansion of: $(x^5+x^6+x^7+\ldots )^2 \cdot (1+x+x^2+ \ldots)^8$
I feel like I have no knowledge that would allow me to even begin to solve this problem. I watched videos about the binomial and multinomial theorems, read the wikipedia articles, and I still have no idea what sort of calculation I'm supposed to do here.
On
As Jean Marie points out in the comments, $$(x^5+\ldots)^2\cdot(1+x+\ldots)^8=x^{10}\cdot(1+x+x^2+\ldots)^{10}$$In other words, we want the coefficient of $x^2$ in $(1+x+x^2+\ldots)^{10}$. Observe that to get a term $ax^2$ from this product, we need the sum of the powers of the terms we are multiplying to be $2$.
More simply put, we are solving the problem $x_1+\ldots+x_{10}=2$, with $x_1,\ldots,x_{10}\in\mathbb N$. The solution to this problem is given by stars and bars to be ${11\choose2}=55$.
You can factorize $x^5$ in the first parenthesis, giving:
$$(x^5)^2(1+x+x^2+\ldots)^2\cdot(1+x+x^2+\ldots)^8=x^{10}\cdot\underbrace{(1+x+x^2+\ldots)^{10}}_P$$
In the expansion of polynomial $P$, we evidently have to look for monomials $ax^2$. Therefore, one can neglect all monomials with degree $>2$ because they cannot contribute to terms in $x^{10}$, i.e., we are interested into the expansion of
$$(1+x+x^2)^{10}=(1+(x+x^2))^{10}$$
which is easily done, purely by binomial theorem (a particular case of multinomial theorem :)):
$$\binom{10}{0}+\binom{10}{1}(x+x^2)+\binom{10}{2}(x+x^2)^2+...=1+10(x+x^2)+45(x+x^2)^2+...$$
We stop there for the same reason as above (no expected contribution to the result)
Therefore the monomial in $x^2$ are $10x^2+45x^2=55x^2$ giving the result $55$.
Remark: if one compares with the method of Don Thousand, our meeting point is that:
$$\binom{10}{0}+\binom{10}{1}=\binom{10}{2}...$$