By Permutation of Indices of Summation: :$\displaystyle \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \sum_{k \mathop = 1}^n \floor {\dfrac {n + 1 - k} 2}$ and so: :$\displaystyle \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }$ 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}}