Summation over k of Floor of k over 2

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Theorem

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


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:

\(\ds \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}\) \(=\) \(\ds \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2\)
\(\ds \) \(=\) \(\ds \dfrac {k - 1 + n + 1 - k} 2\)
\(\ds \) \(=\) \(\ds \dfrac n 2\)


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:

\(\ds \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}\) \(=\) \(\ds \dfrac k 2 + \dfrac {n - k} 2\)
\(\ds \) \(=\) \(\ds \dfrac {k + n - k} 2\)
\(\ds \) \(=\) \(\ds \dfrac n 2\)


So:

\(\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2}\) \(=\) \(\ds \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\dfrac n 2}\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 n \dfrac n 2\)
\(\ds \) \(=\) \(\ds \dfrac {n^2} 4\)
\(\ds \) \(=\) \(\ds \floor {\dfrac {n^2} 4}\) as $\dfrac {n^2} 4$ is an integer

$\Box$


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:

\(\ds \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}\) \(=\) \(\ds \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2 - \dfrac 1 2\)
\(\ds \) \(=\) \(\ds \dfrac {k - 1 + n + 1 - k - 1} 2\)
\(\ds \) \(=\) \(\ds \dfrac {n - 1} 2\)


For $k$ even:

$\floor {\dfrac k 2} = \dfrac k 2$

and:

$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$

Hence:

\(\ds \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}\) \(=\) \(\ds \dfrac k 2 + \dfrac {n - k + 1} 2\)
\(\ds \) \(=\) \(\ds \dfrac {k + n - k + 1} 2\)
\(\ds \) \(=\) \(\ds \dfrac {n + 1} 2\)


Let $n = 2 t + 1$.

Then:

\(\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2}\) \(=\) \(\ds \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \paren {\floor {\dfrac k 2} + \floor {\dfrac {2 t + 2 - k} 2} }\)
\(\ds \) \(=\) \(\ds \dfrac t 2 \dfrac {\paren {2 t + 1} + 1} 2 + \dfrac {t + 1} 2 \dfrac {\paren {2 t + 1} - 1} 2\) there are $t$ even terms and $t + 1$ odd terms
\(\ds \) \(=\) \(\ds \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 2 t} 4\) multiplying out
\(\ds \) \(=\) \(\ds \dfrac {4 t^2 + 4 t} 4 + \dfrac 1 4 - \dfrac 1 4\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {2 t + 1}^2} 4 - \dfrac 1 4\)
\(\ds \) \(=\) \(\ds \dfrac {n^2} 4 - \dfrac 1 4\)
\(\ds \) \(=\) \(\ds \floor {\dfrac {n^2} 4}\)

$\blacksquare$


Sources

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