Prove that $\max\{a_i \mid i \in \{0,1,\ldots,1984\}\} = a_{992}$ 
Consider the expansion
  $$\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496} =
a_{0} + a_{1}x + \cdots + a_{1984}\,\,x^{1984}
$$
  Prove that
  $\max\left\{a_{i} \mid i \in \left\{0,1,\ldots,1984\right\}\right\} =
a_{992}\ $.

Since $1 + x + x^{2} + x^{3} + x^{4}$ is a palindromic polynomial, so is $\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496}$. Now we want to show that $a_{1},a_{2},\ldots,a_{1984}$ has exactly one maximum and we are done.
 A: As you have noticed, for $(1 + x + x^2 + x^3 + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}$, we have
$$
a_i = a_{4n-i}\quad\text{for } 0 \leq i \leq 4n
$$
So it is sufficient to prove that $a_0 < a_1 < \cdots < a_{2n}$. We prove by induction below.

Basis. For $n = 2$, we have
$$
(1 + x + \cdots + x^4)^2 = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 4x^5 + 3x^6 + 2x^7 + x^8
$$
Thus $a_0 < a_1 < \cdots < a_4$.

Induction. Suppose for $n \geq 2$ we have
$$
(1 + x + \cdots + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}
$$
with $a_0 < a_1 < a_2 < \cdots < a_{2n}$ and
$$
(1 + x + \cdots + x^4)^{n+1} = b_0 + b_1x + \cdots + b_{4n+4}x^{4n+4}
$$
We aim to prove that $b_0 < b_1 < \cdots < b_{2n+2}$. Note that
$$
(1 + x + \cdots + x^4)^{n+1} = (1+x+\cdots + x^4)^n\cdot(1+x+\cdots +x^4)
$$
Therefore,
\begin{align}
b_i &= a_i + a_{i-1} + a_{i-2} + a_{i-3} + a_{i-4}\quad \text{for}\quad 4\leq i \leq 2n + 2\\
b_3 &= a_3 + a_2 + a_1 + a_0 \\
b_2 &= a_2 + a_1 + a_0 \\
b_1 &= a_1 + a_0 \\
b_0 &= a_0
\end{align}
Easy to observe that $b_0 < b_1 < b_2 < b_3 < b_4$. For $4 < i \leq 2n$, we have
$$
b_{i} - b_{i-1} = a_i - a_{i-5} > 0
$$
by our induction assumption. Moreover, we have
$$
b_{2n + 1} - b_{2n} = a_{2n+1} - a_{2n-4} = a_{2n-1} - a_{2n-4} > 0
$$
and
$$
b_{2n + 2} - b_{2n+1} = a_{2n + 2} - a_{2n - 3} = a_{2n - 2} - a_{2n - 3} > 0
$$
Therefore,
$$
b_0 < b_1 < \cdots < b_{2n + 2}
$$

When $n = 496$, by our proof, $a_{992}$ is maximum.
A: Let us say that a degree $2n$ polynomial is (SID) (Symmetrical Increasing-decreasing) property if the list of its coefficients $a_0, a_1... a_{2n}$ (considered in ascending power of the indeterminate) is symmetrical ($a_{2n-k}=a_k$) with the first $n$ being ascending, with a unique maximum, then the last $n$ being strictly decreasing.
The property you are aiming at is a simple consequence of the following lemma:
If $P(x)$ is (SID), $Q(x)=P(x)*(1+x+x^2+x^3+x^4)$ is (SID) as well.
An example: $P(x)=(1+x+x^2+x^3+x^4)^2=1+2x+3x^2+4x^3+...$
is (SID) because its coefficients have a tent-shape $(1, 2, 3, 4, 5, 4, 3, 2, 1)$.
then $Q(x)=P(x)*(1+x+x^2+x^3+x^4)=(1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1)$
is also (SID) (looking a little more gaussian).
Proof: Due to the symmetrical property, we only need to work on the first part of the coefficients, i.e. proving that the sequence is ascending and reaches a maximum in its first "half".
The general formula for the coefficients $b_k$ of $Q(x)$ is
$$b_k=a_k+a_{k-1}+a_{k-2}+a_{k-3}+a_{k-4}$$
Explanation : the first coefficients can be considered obeing the same formula by adding a certain  four zeros before and four zeros after the list of coefficients $a_k$ (what is called ''zero-padding'').
Computing $b_{k+1}$ out of $b_k$ amounts to suppress $a_{k-4}$ and introduce $a_{k+1}$, in other words:
$$b_{k+1}=b_k+(a_{k+1}-a_{k-4})$$ 
This formula clearly shows that $b_{k+1}>b_k$ is equivalent to $(a_{k+1}-a_{k-4})>0$, i.e., as long as the maximum hasn't been reached ; more precisely, 


*

*if $k+1=n$, $a_{k+1}-a_{k-4}=a_{n}-a_{n-3}>0$

*if $k+1=n+1$, $a_{k+1}-a_{k-4}=a_{n+1}-a_{n-2}>0$

*if $k+1=n+2$, $a_{k+1}-a_{k-4}=a_{n+2}-a_{n-1}<0$ 
ending the lemma's proof.
Remark : it is like making a moving average : the moving average of a symmetrical, first ascending then descending distribution has the same property.
The final step is the use of the lemma in a recurrent way.
A: In probabilistic terms, if we have a sequence of i.i.d. random variables $X_i$ such that the PDF of $X_i$ is supported on $[0,1]$ and symmetric around $\frac{1}{2}$, the PDF of
$$ S_n=X_1+X_2+\ldots+X_n $$
is supported on $[0,n]$ and symmetric around $\frac{n}{2}$. If $X_1$ is uniformly distributed and $n\geq 2$, the PDF of $S_n$ has a maximum at $\frac{n}{2}$. By the Central Limit Theorem / Berry Esseen theorem, if $n$ is large $S_n$ is well approximated by a normal random variable with mean $\frac{n}{2}$. We are just interested in the discrete analogue of this fact.
