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 $.

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.
