Proof that ordinary multinomial coefficients rise monotonically to a maximum and then decrease monotonically While most computations of ordinary multinomial coefficients for the following case require recursive summations, I found here a closed-form solution:
$$(1+x+x^2+\cdots+x^q)^L = \sum_{a \geq 0} \binom{L}{a}_q x^a,$$
where
$$\binom{L}{a}_q = \sum_{j=0}^{\lfloor a/(q+1) \rfloor} (-1)^j \binom{L}{j} \binom{a-j(q+1)+L-1}{L-1}.$$
with $\binom{L}{a}_1 = \binom{L}{a}$ (being the usual binomial coefficient) and $\binom{L}{a}_q = 0$ for all $a > qL.$
By the nature of multinomial coefficients, and evidenced by generalized Pascal triangles, we know that the coefficients rise monotonically to a maximum located at or near $a=\frac{qL}{2}$ depending on the parity of $qL.$
QUESTION: How do we prove the above statement? I'd like to be able to show, for example, that
$$\binom{L}{a+1}_q / \binom{L}{a}_q > 1 \forall a \in [0,qL/2-1]$$
and similarly
$$\binom{L}{a+1}_q / \binom{L}{a}_q < 1 \forall a \in [qL/2,qL].$$
Ideas? I have tried computing the quotients above directly from the definition of $\binom{L}{a}_q$ above but it is very messy and I am never able to get to a point that I can prove unequivocally the existence and nature of this global maximum.
Thank you in advance for the help!
 A: Call a polynomial $A(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$ unimodal if there is some $t$ such that for $i<j\leq t$, $a_{i} \leq a_{j}$, and for $t \leq i < j$, $a_{i} \geq a_{j}$; in other words, the coefficients "rise monotonically to a maximum and then decrease monotonically". Call the polynomial $A(x)\;$  symmetric if for all $i$, $a_{i} = a_{n-i}$.
We will prove the following lemma: If $A(x)$ and $B(x)$ are two symmetric unimodal polynomials, then $A(x)B(x)$ is a symmetric unimodal polynomial. To see this, first let
$$A(x) = \sum_{i=0}^{m}a_{i}x^{i}$$
$$B(x) = \sum_{j=0}^{n}b_{j}x^{j}$$
Let $r = \lfloor m/2 \rfloor$ and $s = \lfloor n/2 \rfloor$. Then, we can rewrite the above polynomials in the following form:
$$A(x) = \sum_{i=0}^{r}(a_{i}-a_{i-1})(x^{i} + x^{i+1} + \dots + x^{m-i})$$
$$B(x) = \sum_{j=0}^{r}(b_{j}-b_{j-1})(x^{j} + x^{j+1} + \dots + x^{n-j})$$
It follows that
$$A(x)B(x) = \sum_{i=0}^{r}\sum_{j=0}^{s}(a_{i}-a_{i-1})(b_{j}-b_{j-1})(x^{i} + \dots + x^{m-i})(x^{j} + \dots + x^{n-j})$$
Now, note that each term of this sum is a symmetric, unimodal polynomial centered around $x^{(m+n)/2}$. Therefore the sum $A(x)B(x)$ itself is a symmetric, unimodal polynomial centered around $x^{(m+n)/2}$, as desired.
Since the polynomial $(1+x+x^2 + \dots + x^q)$ is symmetric and unimodal, by repeatedly applying this lemma, it follows that $(1+x+x^2 + \dots + x^q)^L$ is symmetric and unimodal, as desired.
Note: unimodal sequences are a well-studied topic in combinatorics. If you're interested in learning more, I recommend the paper "Log-Concave and Unimodal Sequences in Algebra, Combinatorics, and Geometry" by Richard Stanley (can be found here). The lemma and proof above appear as Proposition 1 in that paper, along with many other interesting ways to prove sequences are unimodal.
