TITLE: Asymptotics of $\sum_{k=0}^{n} {\binom n k}^a$
QUESTION [10 upvotes]: I need to estimate the asymptotics of
$$\sum_{k=0}^{n} {\binom n k}^a, \quad  a>2, \quad  a \in \mathbb{N}$$
In particular, I'm pretty much interested in $a=4$ case, but if the general solution isn't very hard to produce and understand, I'd prefer general.
Could you please help me with any method of solving this?
// Added later:
Possibly - could we somehow make use of fact that the most "heavy" members $k$ are located "near" $n/2\quad $ (e.g., $k=n/2 \pm o(\sqrt n)$)?

REPLY [11 votes]: A quick way to get the leading term is to use the fact that ${n \choose k}/2^n$ represents a Binomial(n,1/2), which for large $n$ approximates a normal with $\mu=n/2$ and  $\sigma^2=n/4$
So $$ {\binom n k} \approx 2^{n}  \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left(-\frac{(k-\mu)^2}{2 \sigma^2}\right)  $$
and $$\sum_{k=0}^{n} {\binom n k}^a \approx 2^{a n} \int \left( \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2 \sigma^2}\right)\right)^a dx = $$
$$ = 2^{an} \frac{1}{(2 \pi \sigma^2)^{a/2}} \int \exp\left(-\frac{(x-\mu)^2}{2 \sigma_1^2}\right) dx$$
with $\sigma_1^2=\sigma^2/a = n/(4a)$
So we get
$$ 2^{an} \frac{1}{(2 \pi \sigma^2)^{a/2}} (2 \pi \sigma_1^2)^{1/2} = \frac{2^{an}}{a^{1/2}}  \left( \frac{\pi n}{2}\right)^{(1-a)/2}  $$