probability of even numbers in n tries of coin flip ive came across a problem in my probability class where they ask to find the probability of getting an even number of heads within n tries. They also added the fact that n is even. 
I tried the binomial with Newtons law but i get 1 instead of $1/2$ 
this is what i try to solve 
$$\sum_{k=0}^{n/2} \binom{n}{2k} \left({\frac{1}{2}}\right)^{2k} \left({\frac{1}{2}}\right)^{n-2k}$$
is this good ?
 A: Hint: For the sum $\sum_{k=0}^{n/2} \binom{n}{2k}$, use the binomial theorem to compare $(1+1)^n$ to $(1+(-1))^n$.
A: The first $n-1$ flips in some sense don't matter at all, because the last flip will make the total number of heads even or odd with equal probability. So the probability is $0.5$.
A: Just to elaborate on the slick hint of @Arthur, because the OP seems confused in the comments: Since $(-1)^k=-1$ if $k$ is odd and $(-1)^k=1$ if $k$ is even, you have:
\begin{align}0=(1+(-1))^n=\sum_{k=1}^n(-1)^k\dbinom{n}{k}&=\sum_{k\text{ even}}\dbinom{n}{k}-\sum_{k\text{ odd}}\dbinom{n}{k}\\[0.2cm]2^n=(1+1)^n=\sum_{k=1}^n\dbinom{n}{k}&=\sum_{k\text{ even}}\dbinom{n}{k}+\sum_{k\text{ odd}}\dbinom{n}{k}\end{align} Hence, adding these two you get: $$2^n+0=\sum_{k\text{ even}}\dbinom{n}{k}+\sum_{k\text{ odd}}\dbinom{n}{k}+\sum_{k\text{ even}}\dbinom{n}{k}-\sum_{k\text{ odd}}\dbinom{n}{k}=2\sum_{k\text{ even}}\dbinom{n}{k}$$ which implies that $$2^n=2\sum_{k\text{ even}}\dbinom{n}{k} \implies \sum_{k\text{ even}}\dbinom{n}{k}=2^{n-1}$$ Hence $$\frac1{2^n}\sum_{k=0}^{n/2}\dbinom{n}{2k}=\frac1{2^n}\sum_{k\text{ even}}\dbinom{n}{k}=\frac{1}{2^n}2^{n-1}=\frac12$$
