TITLE: Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$
QUESTION [10 upvotes]: Can this sum be simplified: $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$
Or at least is there a simple fairly tight upperbound?
EDIT
So I think this sum is more easily bounded than I previously thought:
Clearly, $$\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \cdot \leq 2^{\sqrt{n}} \sum_{k=0}^{n} \binom{n}{k} = 2^{n+\sqrt{n}} .$$
Also, along the same lines as Shai Covo's answer, $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \geq \binom{n}{n/2} 2^{\sqrt{n/2}}$. The central binomial coefficient: $\binom{n}{n/2}$ is at least $\frac{2^n}{\sqrt{2n}}$, hence $$\sum_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}} \geq 2^{\sqrt{n/2}} \cdot \frac{2^n}{\sqrt{2n}} = \frac{2^{n + \sqrt{n/2}}}{2n}$$
So $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ is $O(2^{n + \sqrt{n}})$ and $\Omega(\frac{2^{n+\sqrt{n/2}}}{n})$
What about expressions of the form:
$\sum \limits_{k=0}^{n} \binom{n}{k} a^{\sqrt{k}} b^{\sqrt{n-k}}$?

REPLY [2 votes]: The mass in the sum is approximately proportional to a (shifted) Gaussian centered at $k = n/2 + C\sqrt{n}$ and with standard deviation $A \sqrt{n}$ for explicitly calculable constants $C$ and $A$.  This implies asymptotics of the form $(M + o(1))2^{n + \sqrt{n/2}}$ where $M$ is another computable constant. 
The same is true for the $a,b$ version of the problem.
[edit: my calculations give $M = 2^{(\ln 2)/8} = 1.0618966...$.
[edit-2: similar calculations give $M=(a/b)^{\ln (a/b)}/8$ and asymptotics $(M+o(1))2^n (ab)^\sqrt{n/2}$ for the $a,b$ sum.]