Proving a combinatorial identity $\sum_{i = 0}^{[n/2]} \binom{n}{2i} p^{2i}q^{n-2i} = \frac{1}{2}\left[(p+q)^n + (q-p)^n \right]$ I would really appreciate some help with proving following identity. I've been trying for the whole day to no avail.
$$\sum_{i = 0}^{[n/2]} \binom{n}{2i} p^{2i}q^{n-2i} = \frac{1}{2}\left[(p+q)^n + (q-p)^n \right]$$
Where by$\ [n/2]$ we mean the largest integer less than or equal to $\ n/2 $. Also $\ q = 1 - p$ 
This question is a step in a probability problem that i'm solving hence the $\ q = 1 - p $.
 A: Expand the righthand side:
$$\begin{align*}
\frac12\left((p+q)^n+(-p+q)^n\right)&=\frac12\left(\sum_{k=0}^n\binom{n}kp^kq^{n-k}+\sum_{k=0}^n\binom{n}k(-1)^kp^kq^{n-k}\right)\\\\
&=\frac12\sum_{k=0}^n\binom{n}k\left(1+(-1)^k\right)p^kq^{n-k}\\\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k2p^{2k}q^{n-2k}\\\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}kp^{2k}q^{n-2k}\;,
\end{align*}$$
since $$1+(-1)^k=\begin{cases}2,&\text{if }k\text{ is even}\\0,&\text{if }k\text{ is odd}\;.\end{cases}$$
A: From 
$$(q\pm p)^n=\sum_{j=0}^n{n\choose j}(\pm 1)^{j}p^jq^{n-j} $$
we see that adding both sums cancels all terms with odd $j$. After halving the sum and replacing even $j$ with $2i$, we obtain the desired result.
A: Suppose we don't have the RHS. Write:
$$
f(z) = \sum_k \binom{n}{k} p^{n - k} z^k = (p + z)^n
$$
We essentially want the even terms of this for $z = q$. For any series $f(z)$ you get the even terms by $(f(z) + f(-z)) / 2$ (and odd ones through $(f(z) - f(-z)) / 2$):
$\begin{align}
\sum_i \binom{n}{2 i} p^{n - 2 i} z^{2 i} 
 &= \frac{f(z) + f(-z)}{2} \\
 &= \frac{1}{2} \left( (p + z)^n + (p - z)^n \right)
\end{align}$
With $z = q$ you get the claim.
