Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$. 
Is there an easy way to obtain this closed-form?
Also, if there are any good references for binomial coefficient identities like these I'd appreciate it. I searched some but did not find any similar to this.

Permit me to contribute a proof complex variables, for variety's sake,
which is an instructive exercise.
Suppose we seek to verify that
$$\sum_{q=0}^{\lfloor n/2 \rfloor} 
{n-q\choose q} 2^q
= \frac{1}{3} \left(2^{n+1} + (-1)^n\right)$$
with $n$ a positive integer.
Introduce the integral representation
$${n-q\choose q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-q}}{z^{q+1}} \; dz$$
Note that  this zero for  $\lfloor n/2 \rfloor  \lt q\le n$ so  we may
extend the sum to $n,$ getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\sum_{q=0}^n
\frac{2^q}{(1+z)^q z^{q}} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\frac{2^{n+1}/(1+z)^{n+1}/z^{n+1}-1}
{2/(1+z)/z-1} \;dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\frac{2^{n+1}/(1+z)^{n}/z^{n}-z(1+z)}
{2-z(1+z)} \;dz.$$
This has two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{2^{n+1}}{2-z(1+z)} \;dz
\\ = \frac{2^{n+1}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(\frac{1}{3}\frac{1}{1-z} + \frac{1}{6}\frac{1}{1+z/2}\right)
\; dz.$$
Extracting coefficients we find
$$2^{n+1}
\left(\frac{1}{3} + \frac{1}{6}\frac{(-1)^n}{2^n}\right)
= \frac{1}{3}
\left(2^{n+1} + (-1)^n\right).$$
The second piece is
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n+1}
\frac{1}{2-z(1+z)} \;dz$$
which gives a contribution of zero.

Collecting everything we obtain
$$\frac{1}{3}
\left(2^{n+1} + (-1)^n\right).$$
