By Permutation of Indices of Summation: :$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \sum_{k \mathop = 1}^n \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil$ and so: :$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)$ First take the case where $n$ is even. For $k$ odd: :$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2 + \dfrac 1 2$ and: :$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2$ Hence: {{begin-eqn}} {{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil       | 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} 2       | c =  }} {{end-eqn}} For $k$ even: :$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2$ and: :$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2 + \dfrac 1 2 = \dfrac {n - k + 2} 2$ Hence: {{begin-eqn}} {{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil       | r = \dfrac k 2 + \dfrac {n - k + 2} 2       | c =  }} {{eqn | r = \dfrac {k + n - k + 2} 2       | c =  }} {{eqn | r = \dfrac {n + 2} 2       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil       | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)       | c =  }} {{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\dfrac {n + 2} 2}\right)       | c =  }} {{eqn | r = \dfrac 1 2 n \left({\dfrac {n + 2} 2}\right)       | c =  }} {{eqn | r = \dfrac {n \left({n + 2}\right)} 4       | c =  }} {{eqn | r = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil       | c = as $\dfrac {n \left({n + 2}\right)} 4$ is an integer }} {{end-eqn}} {{qed|lemma}} Next take the case where $n$ is odd. For $k$ odd: :$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2 + \dfrac 1 2$ and: :$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2 + \dfrac 1 2$ Hence: {{begin-eqn}} {{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil       | 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 + 3} 2       | c =  }} {{end-eqn}} For $k$ even: :$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2$ and: :$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2$ Hence: {{begin-eqn}} {{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil       | 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 \left \lceil{\dfrac k 2}\right \rceil       | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)       | c =  }} {{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {2 t + 2 - k} 2}\right \rceil}\right)       | c =  }} {{eqn | r = \dfrac t 2 \dfrac {\left({2 t + 1}\right) + 1} 2 + \dfrac {t + 1} 2 \dfrac {\left({2 t + 1}\right) + 3} 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 + 6 t + 4} 4       | c = multiplying out  }} {{eqn | r = \dfrac {4 t^2 + 8 t + 3} 4 + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {\left({2 t + 1}\right) \left({2 t + 3}\right)} + \dfrac 1 4       | c =  }} {{eqn | r = \dfrac {n \left({n + 2}\right)} 4 + \dfrac 1 4       | c =  }} {{eqn | r = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil       | c =  }} {{end-eqn}} {{qed}}