If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$ 
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$ 
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up: 
$$=.64$$ so there's a 64% chance of probability?

Nine coins, so that two events 
$\mathscr{E}_1$ = #heads is even and 
$\mathscr{E}_2$ = #tails is even
are mutually exclusive (the number of tails is 9 - number of heads, so former is even iff latter odd) and comprise all possibilities, thus $P(\mathscr{E}_1) + P(\mathscr{E}_2) =1$. But if the coins are fair, then the probabilities must be unchanged if we swap the roles of heads and tails. Hence $P(\mathscr{E}_1)= P(\mathscr{E}_2)$ and we immediately see both probabilities must be $\frac{1}{2}$.

Now you are wondering why your approach doesn't work, because it is basically sound. You've simply made a slip. 
You're approach is: sum every second term in the 10 member (i.e. an even number of terms) sequence whose $n^{th}$ term is the probability of $n$ heads. So the sum is:
$$S_1=\sum_{k=0}^{N/2} \binom{N}{2\,k}\left(\frac{1}{2}\right)^N$$
with $N$ odd (here equal to 9).
But, by dint of $\binom{N}{2\,k} = \binom{N}{N-2\,k}$, this sum is equal to the sum of all the other terms
$$S_2 =\sum_{k=0}^{N/2} \binom{N}{N-2\,k}\left(\frac{1}{2}\right)^N$$
in the sequence that don't belong to the first sum. So $S_1=S_2$ and clearly $S_1+S_2=1$, because this sum is the sum of probabilities of all possible mutually exclusive outcomes, therefore 1, or, alternatively, call up the binomial theorem and see that $S_1+S_2=\left(\frac{1}{2} + \frac{1}{2}\right)^9=1$
