Theorem: Summation over k of Floor of k over 2

:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \floor {\dfrac {n^2} 4}$

Proof:
By Permutation of Indices of Summation:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \sum_{k \mathop = 1}^n \floor {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }$
First take the case where $n$ is even.
For $k$ odd:
:$\floor {\dfrac k 2} = \dfrac k 2 - \dfrac 1 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
      | r = \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2
      | c = 
}}
{{eqn | r = \dfrac {k - 1 + n + 1 - k} 2
      | c = 
}}
{{eqn | r = \dfrac n 2
      | c = 
}}
{{end-eqn}}
For $k$ even:
:$\floor {\dfrac k 2} = \dfrac k 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 - \dfrac 1 2 = \dfrac {n - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
      | r = \dfrac k 2 + \dfrac {n - k} 2
      | c = 
}}
{{eqn | r = \dfrac {k + n - k} 2
      | c = 
}}
{{eqn | r = \dfrac n 2
      | c = 
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \floor {\dfrac k 2}
      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }
      | c = 
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\dfrac n 2}
      | c = 
}}
{{eqn | r = \dfrac 1 2 n \dfrac n 2
      | c = 
}}
{{eqn | r = \dfrac {n^2} 4
      | c = 
}}
{{eqn | r = \floor {\dfrac {n^2} 4}
      | c = as $\dfrac {n^2} 4$ is an integer
}}
{{end-eqn}}
{{qed|lemma}}
Next take the case where $n$ is odd.
For $k$ odd:
:$\floor {\dfrac k 2} = \dfrac k 2 - \dfrac 1 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 - \dfrac 1 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
      | r = \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2 - \dfrac 1 2
      | c = 
}}
{{eqn | r = \dfrac {k - 1 + n + 1 - k - 1} 2
      | c = 
}}
{{eqn | r = \dfrac {n - 1} 2
      | c = 
}}
{{end-eqn}}
For $k$ even:
:$\floor {\dfrac k 2} = \dfrac k 2$
and:
:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
      | r = \dfrac k 2 + \dfrac {n - k + 1} 2
      | c = 
}}
{{eqn | r = \dfrac {k + n - k + 1} 2
      | c = 
}}
{{eqn | r = \dfrac {n + 1} 2
      | c = 
}}
{{end-eqn}}
Let $n = 2 t + 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \floor {\dfrac k 2}
      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }
      | c = 
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \paren {\floor {\dfrac k 2} + \floor {\dfrac {2 t + 2 - k} 2} }
      | c = 
}}
{{eqn | r = \dfrac t 2 \dfrac {\paren {2 t + 1} + 1} 2 + \dfrac {t + 1} 2 \dfrac {\paren {2 t + 1} - 1} 2
      | c = there are $t$ even terms and $t + 1$ odd terms 
}}
{{eqn | r = \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 2 t} 4
      | c = multiplying out 
}}
{{eqn | r = \dfrac {4 t^2 + 4 t} 4 + \dfrac 1 4 -  \dfrac 1 4
      | c = 
}}
{{eqn | r = \dfrac {\paren {2 t + 1}^2} 4 - \dfrac 1 4
      | c = 
}}
{{eqn | r = \dfrac {n^2} 4 - \dfrac 1 4
      | c = 
}}
{{eqn | r = \floor {\dfrac {n^2} 4}
      | c = 
}}
{{end-eqn}}
{{qed}}