Odd Power terms of binomial theorem proof I want to acquire all the terms of $(p+q)^n$ where the power of p is odd. 
Note that $p=1-q$ ($p$,$q$ probabilities)
Ex. For $(p+q)^2=p^2+q^2+2pq$  I want to acquire only $2pq$(only term with odd power of p)
I know that the relation I am looking is the following
$$\frac{(q+p)^n-(q-p)^n}2$$
My question is how can I prove that this is the case?? Is there an identity or something that can lead to a proof?

$$(p+q)^n=\sum_{i=0}^n\binom{n}{i}p^{n-i} q^{i}$$
$$(p-q)^n=\sum_{i=0}^n(-1)^i\binom{n}{i}p^{n-i} q^{i}$$
Differens of the two
$$(p+q)^n-(p-q)^n=\sum_{i=0}^n\binom{n}{i}p^{n-i} q^{i}-\sum_{i=0}^n(-1)^i\binom{n}{i}p^{n-i} q^{i}$$
$$=\sum_{i=0}^n\binom{n}{i}p^{n-i} q^{i}-(-1)^i\binom{n}{i}p^{n-i} q^{i}$$
$$=\sum_{i=0}^n\binom{n}{i}p^{n-i} q^{i}(1-(-1)^i)$$
for even $n$ we have $(-1)^i=1$ thereby $(1-1)=0$ and for odd ones we have $(-1)^i=-1$ so $(1-1)=2$ which gives us
$$2\sum_{0\leq i\leq n,2\nmid i}\binom{n}{i}p^{n-i} q^{i}$$
The rest follows
