If $n$ is odd, then $n/2 + 1/2$ is always even? 
If $n$ is odd, prove that $n/2 + 1/2$ is even.

Context: I'm a Statistician and the term $n/2 + 1/2$ showed up in the index of a summation when deriving the pdf of some Order Statistic:
$$
\sum_{j = (n+1)/2}^{n}...
$$
I realized that $n/2 + 1/2$ is always even if $n$ is odd, but I couldn't prove the result to myself (well, I have no training in Number Theory).
What I've tried:
Suppose $n$ is odd. Then $n + 1$ is even (by the successor function?). Then $n + 1 = 2k$ for $k \in \mathbb{N} \Rightarrow (n+1)/2 = k $. But this doesn't show that $k$ is even.

If $n$ is odd, then $n=2k+1$ for some integer $k$. So $$\frac{n+1}{2} = \frac{2k+1+1}{2} = k+1.$$So your statement is false. E.g., for $n=1$ we have $(n+1)/2=1$ odd and for $n=3$ we have $(n+1)/2=2$ even.
