Largest coefficient in the power of a polynomial

For something like $(1+2x+2x^2)^n$, you can say that the expected power of $x$ for each factor is $\frac 65$, so you would expect the maximum to be at $\frac 65n$. For $n=100$ this would be the $x^{120}$ term and it truly is per Alpha. You have to click more terms a bunch to see this. It is basically using the normal approximation to the binomial distribution.

1. Write

$$\left(\frac{1}{5} + \frac{2x}{5} + \frac{2x^2}{5}\right)^n = \sum_{l=0}^{2n} a_lx^l.$$

2. Let $x_1,x_2,\ldots, x_n$ be iids where each $x_i$ is 0 with probability $\frac{1}{5}$; 1 with probability $\frac{2}{5}$; and 2 with probability $\frac{2}{5}$. Then for each $l$, note the following:

$${\bf{P}}\left[\left(\sum_{i=1}^n x_i \right) = l \right] = a_l$$

3. It turns out that (Chernoff bounds) that the value of $l$ that maximizes $a_l={\bf{P}}\left[\left(\sum_{i=1}^n x_i \right) = l \right]$ is $l \approx {\bf{E}}\left[\sum_{i=1}^n x_i \right]$ which is $l \approx \frac{6n}{5}$.

4. So putting the above together, writing $\left(\frac{1}{5} + \frac{2x}{5} + \frac{2x^2}{5}\right)^n$ as $\sum_{l=0}^{2n} a_lx^l$, the value of $a_l$ such that $a_l$ is the largest is $l \approx \frac{6n}{5}$ and for such $l$, the coefficient $a_l$ has value $\theta \left(\frac{1}{\sqrt{n}}\right)$.

5. So writing

$$(1+2x+2x^2) = \sum_{l=0}^{2n} b_lx^l,$$

the value of $l$ that maximzes $b_l$ is $l\approx \frac{6n}{5}$, and $b_l$ has value $\theta(\frac{5^n}{\sqrt{n}})$.