Largest coefficient in the power of a polynomial How do I find the largest coefficient of any power of $x$ in an expansion such as $(1 + 2x + 2x^2)^n$, as a function of $n$?
In the case $(1+x)^n$, we know that the central coefficients are the largest, and for $(1+ax)^n$ I can take ratios of consecutive terms to find the largest one, but these methods fail in the case mentioned above.
Also, can we find the degree of the term for which this largest coefficient occurs, again as a function of $n$?
 A: 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.

A: *

*Write 


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


*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:


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


*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 $l \approx \frac{6n}{5}$.

*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 $\theta \left(\frac{1}{\sqrt{n}}\right)$.

*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  $\theta(\frac{5^n}{\sqrt{n}})$.
