Asymptotic approximation of the expression $\sum _{i=0}^n \log (\binom{n}{i}+1)$ I am trying to work out the asymptotic approximation (for large $n$) of the following:
 $$\sum _{i=0}^n \log (\binom{n}{i}+1)$$
So far I tried to write the expression as, 
$$\sum^{n}_{i=0}log(x)=-\sum^{n}_{i=0}x\int^{\infty}_{0}e^{-xt}\log tdt-\gamma$$
where I took $x=\binom{n}{i}+1$ and $\gamma$ is Euler gamma, but I am not sure if this method is so useful. Any other approach is appreciated.  
 A: The proposer has pointed out the dominant term and the addition of 1 within the log argument is a minor correction.
$$S:=\sum_{k=0}^{n}\log(1+\binom{n}{k}) =\sum_{k=0}^{n} \log(\binom{n}{k}(1+1/{\binom{n}{k}})\sim$$
$$\sim\sum_{k=0}^{n} \log(\binom{n}{k})+\sum_{k=0}^n\frac{1}{\binom{n}{k}}-\frac{1}{2}\sum_{k=0}^{n}\frac{1}{\binom{n}{k}^2}+...$$
where a Taylor series of $\log(1+x)$ has been used.
It can be shown that
$$\sum_{k=0}^n\frac{1}{\binom{n}{k}} =2(1+\frac{1}{n}+...) \text{ and } \sum_{k=0}^n\frac{1}{\binom{n}{k}^2}=2(1+1/n^2+...)$$
We will always know the leading coefficient is 2 because of the $k=0$ and $k=n$ terms.  We have an infinite sum of these terms, which fortunately converges because of their alternating weight of $(-1)^m/m.$ The log-binomial sum can be done in closed form:
$$\sum_{k=0}^{n} \log(\binom{n}{k})=\log\Big(\frac{n!^{n+1}}{G(n+2)^2}\Big). $$
$G$ is the Barne's G and its asymptotic expansion is well-known.
Collecting, 
$$S\sim \log\Big(\frac{n!^{n+1}}{G(n+2)^2}\Big) + 2\log{2}+\frac{2}{n}+O(\frac{1}{n^2}) $$
Use this expansion for $n=20$ and 4 correct digits are obtained.
