How many binary words of length n , that consist an even number of zeros? How many binary words (chars '0' and or '1') of length n that consist an even number of zeros are there?
I know that there are $2^n$ options overall, and that for every $n$, there are $\lceil{\frac{n}{2}}\rceil + 1$ options for even zeros. But now what? I got lost! 
Maybe I need to use the pascal triangle?
 A: It should be $\lfloor{\frac{n}{2}}\rfloor + 1$ (try it with $n=2,3$), but you don't need that.  Take any option for the first $n-1$ bits, then the last is determined by the parity.
A: Doesn't common sense say that there should be an equal number of strings with an even number of zeros as with an odd number of zeros?  Well, not if n is odd I guess.  But in that case shouldn't common sense say there are just as many strings with an even number of zeros as there are with an even number of ones (which would have an odd number of zeros).
But we shouldn't rely on common sense.  But we shouldn't toss it out either.
==== common sense formalized: half the strings have an even number of zeros, and half have an odd number of zeros ===========
Let $a = [a_i]$ be an n digit binary number.  (Each $a_i$ is a 0, 1 digit).
Let $f(a) = [b_0a_1... a_i...]$ where the first digit is changed from a 0 to a 1 or from a 1 to a 0.  The rest of the digits are left the same.  
$f$ is clearly a bijection.  $f(a)$ will have either 1 more zero or 1 less zero than $a$.  So if $a$ has an even/odd number of zeros $f(a)$ will have an odd/even number.  So $f$ allows for a 1-1 correspondence between numbers with even number of zeros and numbers with odd number of zeros.
So the number of numbers with even number of zeros is half the total number.
There are $2^{n-1}$ binary strings with an even number of zeros.
A: Hints:
Let $a_n$ be the number that you are looking for.
If $n$ is odd, there are as many words with an even number of zeros as many words with an odd number of zeros, because the complementary words of the former are exactly the latter. 
If $n$ is even, then you can split the words in half, and if the word has an even number of zeros then both halves have an odd number of zeros or both even, so $a_n=2a_{n/2}^2$.
