How can I calculate $ \sum\limits_{j=0}^{49}\binom{100}{2j+1}p^{100-(2j+1)} q^{2j+1} $?

Question

I got the following formula when I tried an exercise in probability: $$ \sum_{j=0}^{49}\binom{100}{2j+1}p^{100-(2j+1)} q^{(2j+1)} $$ where $p+q=1$. These are the "odd" terms in the expansion of $1=(p+q)^{100}$. I guess it might be $1/2$, but I don't have a proof. More generally, how can one simplify $$ \sum_{j=0}^{(n/2)-1}\binom{n}{2j+1}p^{n-(2j+1)} q^{(2j+1)} $$ when $n$ is even?

Answer 1

Note that $$(p+q)^n=\sum_k{n\choose k}p^{n-k}q^k,$$ and $$(p-q)^n=\sum_k(-1)^k{n\choose k}p^{n-k}q^k,$$ hence $$1-(p-q)^n=(p+q)^n-(p-q)^n=2\sum_{k\ \text{odd}}\cdots$$