Summation over k of Floor of k over 2

Theorem

 * $\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\dfrac k 2}\right \rfloor = \left \lfloor{\dfrac {n^2} 4}\right \rfloor$

Proof
By Permutation of Indices of Summation:
 * $\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\dfrac k 2}\right \rfloor = \sum_{k \mathop = 1}^n \left \lfloor{\dfrac {n + 1 - k} 2}\right \rfloor$

and so:
 * $\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\dfrac k 2}\right \rfloor = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lfloor{\dfrac k 2}\right \rfloor + \left \lfloor{\dfrac {n + 1 - k} 2}\right \rfloor}\right)$

First take the case where $n$ is even.

For $k$ odd:
 * $\left \lfloor{\dfrac k 2}\right \rfloor = \dfrac k 2 - \dfrac 1 2$

and:
 * $\left \lfloor{\dfrac {n + 1 - k} 2}\right \rfloor = \dfrac {n + 1 - k} 2$

Hence:

For $k$ even:
 * $\left \lfloor{\dfrac k 2}\right \rfloor = \dfrac k 2$

and:
 * $\left \lfloor{\dfrac {n + 1 - k} 2}\right \rfloor = \dfrac {n + 1 - k} 2 - \dfrac 1 2 = \dfrac {n - k} 2$

Hence:

So:

Next take the case where $n$ is odd.

For $k$ odd:
 * $\left \lfloor{\dfrac k 2}\right \rfloor = \dfrac k 2 - \dfrac 1 2$

and:
 * $\left \lfloor{\dfrac {n + 1 - k} 2}\right \rfloor = \dfrac {n + 1 - k} 2 - \dfrac 1 2$

Hence:

For $k$ even:
 * $\left \lfloor{\dfrac k 2}\right \rfloor = \dfrac k 2$

and:
 * $\left \lfloor{\dfrac {n + 1 - k} 2}\right \rfloor = \dfrac {n + 1 - k} 2$

Hence:

Let $n = 2 t + 1$.

Then: