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., $k=n/2 \pm o(\sqrt n)$)?

Edit: Now in beautiful high-definition technicolor!
$$\sum \limits_{k=1}^n \sqrt{\color{red}1+\color{Green}{\frac{1}{k^2}}+\color{Blue}{\frac{1}{(k+1)^2}}}$$
$$=\sum_{k=1}^n\sqrt{\frac{\color{red}{k^2(k+1)^2}+\color{Green}{(k+1)^2}+\color{Blue}{k^2}}{k^2(k+1)^2}}$$
$$=\sum_{l=1}^n \sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}}$$
$$=\sum_{k=1}^n\sqrt{\frac{\color{Red}1k^4+(\color{Red}1+\color{Green}1)k^3+(\color{Red}1+\color{Green}1+\color{Blue}1)k^2+(\color{Green}1+\color{Blue}1)k+\color{Blue}1}{k^2(k+1)^2}} $$
$$=\sum_{k=1}^n\sqrt{\frac{\color{Red}{k^2(k^2+k+1)}+\color{Green}{k(k^2+k+1)}+\color{Blue}{(k^2+k+1)}}{k^2(k+1)^2}} $$

$$=\sum_{k=1}^n\sqrt{\frac{(k^2+k+1)^2}{k^2(k+1)^2}}$$
$$=\sum_{k=1}^n\frac{\color{Purple}{k^2+k}+\color{Blue}1}{k^2+k}$$
$$=\sum_{k=1}^n \left(\color{Purple}1+\color{Blue}{\frac{1}{k(k+1)}}\right) $$
$$=\sum_{k=1}^n \left(1+\frac{1}{k}-\frac{1}{k+1}\right)$$ 
$$=n+1-\frac{1}{n+1}.$$
Specializing $n=69$ gives $69+1-1/70=4899/70$. Note that it's highly useful to memorize this:
$$\frac{1}{k(k+a)}=\frac{1}{a}\frac{(k+a)-k}{k(k+a)}=\frac{1}{a}\left(\frac{1}{k}-\frac{1}{k+a}\right).$$
