<s>[INST] 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.
 [/INST] If $n$ is odd then $n$ is of the form $2k - 1$.
The number $\frac n2 +\frac 12$ is simply $\frac {n+1}2 = \frac {2k-1 +1}2 = \frac {2k}2 = k$.  The expression $\frac n2 + \frac 12$ is simply the the index of which odd number $n$ is.
If $n = 1,3,5,7,9,......, 2k -1 ,.....$ then $\frac n2 + \frac 12 = 1,2,3,4,5,.....,k,.....$ and $k$ is even only if $n$ is the in an even position on the list and odd if $n$ is in an odd position on the list.
....
There are $2$ types of odd numbers.  Those where $n = 2k-1$ and $k$ is even, and those where $n=2k-1$ and $k$ is odd.
If $k = 2m$ is even then $n= 2(2m)-1 = 4m -1$ and $\frac n2 +\frac 1n = \frac {2k-1}2 + \frac 12 = k = 2m$ is even.
If $k = 2m -1$ then $n = 2(2m-1) -1 = 4m -3$ and $\frac n2 + \frac 1n = \frac {2k-1}2 +\frac 12= k = 2m-1$ is odd.
So $k$ will be even if $n$ is of the form $4m -1$ or $4m+3$ so for example if $n = 3,7,11,15,....$ then $\frac n2 + \frac 12$ will be even.
But $k$ will be odd if $n$ is of the form $4m -3$ or $4m + 1$ so for example if $n=1,5,9,13,17,..$ then $\frac n2 + \frac 12$ will be even.
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