Is there a nice way of expanding multinomials of the form $(1+x+\cdots+x^l)^n$? I have found an expression for the probability generating function of a random variable:
$G_Y(s)=\frac{1}{(6-k+1)^3}[s^k+\cdots+s^6]^3$
(where $k\in\lbrace1,..,6\rbrace$) and I don't know how to use the multinomial theorem to expand this into something usable. 
Essentially I am looking for a formula for the coefficient of $x^i$ in:
$(1+x+\cdots+x^l)^n$
with an explanation being even better. Thanks
 A: 
Essentially I am looking for a formula for the coefficient of $x^i$ in:
$(1+x+\cdots+x^l)^n$

$(1+x+\cdots+x^l)^n = \frac{(1-x^{l+1})^n}{(1-x)^n} = (\sum_{k\ge0} \binom{n-1+k}{n-1}x^k)(\sum_{k=0}^{n} \binom{n}{k}(-x)^{(l+1)k})$
The coefficient of $x^i$ in this expression is $\sum_{p=0}^{n} \binom{n-1+i-(l+1)p}{n-1} \binom{n}{p}(-1)^p$
A: This is a tricky one, in general. The value is $\left(\frac{1-x^{l+1}}{1-x}\right)^n$. Since $\frac{1}{(1-x)^n}=\sum_{k=0}^{\infty} \binom{k+n-1}{n-1}x^n$ and $(1-x^{l+1})^n = \sum_{k=0}^{n} (-1)^k\binom{n}{k}x^{(l+1)k}$, you can get the coefficient for $x^m$ as:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{m-(l+1)k+n-1}{n-1}$$
That's not a lovely formula, but I believe it is the best you can do. 
This sum can also be seen as an inclusion-exclusion formula. If $A$ is the set of all $n$-tuples of non-negative integers $(a_1,a_2,\dots,a_n)$ so that $a_1+\cdots + a_n=m$, and $A_i$ is the subset where $a_i>l$, then you want:
$$|A\setminus (A_1\cup A_2\cup\cdots\cup A_n)|$$
which, when inclusion-exclusion is applied:
$$=|A|-(|A_1|+|A_2|+\cdots+|A_n|) + (|A_1\cap A_2|+|A_1\cap A_3|+\cdots + |A_{n-1}\cap A_n|)-\cdots$$ yields this same formula.
A: Let's start with an easier problem:  What is the coefficient of $x^i$ in 
$ \left( \frac{1}{1-x} \right)^n$ (which we write as $ [x^i]\left( \frac{1}{1-x} \right)^n$)?
Here, each of $n$ factors have the form $\sum_0^\infty x^\ell$ so in order to obtain a term of the form $x^i$ we must merely partition $i$ into an sum $m_1 + m_2 + \cdots + m_n = i$. This is easy if you consider it is isomorphic to placing $n-1$ walls amongst $i$ objects, which in turn is the same as choosing $n-1$ positions for the walls out of $n-1+i$ wall-or-x things. So
$$
[x^i]\left( \frac{1}{1-x} \right)^n = \binom{n+i-1}{i-1}
$$
Say we wanted $[x^i]\left( \frac{1}{1-x} \right)^n x^r $. That would be the same as $[x^{i-r}]\left( \frac{1}{1-x} \right)^n $ which is
$$[x^i]\left( \frac{1}{1-x} \right)^n x^r=\binom{n+i-r-1}{i-r-1}$$.
Now we relate this to the original problem:
$$
\left(1+x+\cdots+x^l\right)^n = \left( \frac{1-x^{l+1}}{1-x}\right)^n 
= \sum_k \left( (-1)^k\frac{\binom{n}{k}x^{k(l+1)}}{(1-x)^n}\right)\\
 [x^i]\left(1+x+\cdots+x^l\right)^n = \sum_k (-1)^k\binom{n}{k}\binom{n+i-k(l+1)-1}{i-k(l+1)-1}
$$
For modest particular values of $\ell$ and $n$ you can express this sum in closed form; for the general case, I don't think you can. At least, Mathematica, which tryes Gosper's algorithm to sum binomials, cannot express this in closed form.
