<s>[INST] 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.
 [/INST] Generating functions to the rescue!
\begin{align*}
\sum_{\substack{n\ge 0 \\ n \text{ even}}} x^n \sum_{i=0}^{n/2} \binom{n-i}{i}2^i &= \sum_{i=0}^\infty 2^i \sum_{\substack{n\ge 2i \\ n \text{ even}}} \binom{n-i}{i} x^n \\
&= \sum_{i=0}^\infty 2^i \sum_{\substack{m\ge 0 \\ m \text{ even}}} \binom{m+i}{i} x^{m+2i} \\
&= \sum_{i=0}^\infty 2^i x^{2i} \sum_{m\ge0} \frac{1+(-1)^m}2 \binom{m+i}{i} x^m \\
&= \frac12 \sum_{i=0}^\infty (2x^2)^i \sum_{m\ge0} \binom{m+i}{i} \big( x^m + (-x)^m \big).
\end{align*}
Using the known series $\sum_{m\ge0} \binom{m+i}{i} x^m = (1-x)^{-(i+1)}$, we get
\begin{align*}
\sum_{\substack{n\ge 0 \\ n \text{ even}}} x^n \sum_{i=0}^{n/2} \binom{n-i}{i}2^i &= \frac12 \sum_{i=0}^\infty (2x^2)^i \big( (1-x)^{-(i+1)} + (1+x)^{-(i+1)} \big) \\
&= \frac1{2(1-x)} \sum_{i=0}^\infty \bigg( \frac{2x^2}{1-x} \bigg)^i + \frac1{2(1+x)} \sum_{i=0}^\infty \bigg( \frac{2x^2}{1+x} \bigg)^i \\
&= \frac1{2(1-x)} \frac1{1-2x^2/(1-x)} + \frac1{2(1+x)} \frac1{1-2x^2/(1+x)} \\
&= \frac{1-2x^2}{1-5x^2+4x^4} \\
&= \frac{2}{3 (1-4x^2)}+\frac{1}{3(1-x^2)} \\
&= \sum_{m=0}^\infty \bigg( \frac23 (4x^2)^m + \frac13 (x^2)^m \bigg) \\
&= \sum_{\substack{n\ge0 \\ n\text{ even}}} \frac{2^{n+1} + 1}3 x^n.
\end{align*}
Comparing coefficients of $x^n$ on both sides establishes the result (without knowing what the answer was in advance, for that matter). Note that $(-1)^n=1$ for $n$ even, so we really got the right answer.
</s>