Endliche Summen(2)

\begin{equation} \sum\limits_{k=0}^{n}{\left\lfloor {\frac{k}{p}} \right\rfloor} = \frac{1}{2} {\left\lfloor {\frac{n}{p}} \right\rfloor} {\left({ 2\,(n+1)\, – p{{\left\lfloor {\frac{n}{p}} \right\rfloor} – p }}\right)}, \quad p \in \mathbb{N} \end{equation}

\begin{align} \sum\limits_{k=0}^{n}{(-1)^k\left\lfloor {\frac{k}{p}} \right\rfloor} =  &\frac{1}{8} {\left[ 2 \left(1 + (-1)^p+2(-1)^n \right) {\left\lfloor {\frac{n}{p}} \right\rfloor} – {\left( 1-(-1)^p \right)} {\left( 1-(-1)^{\left\lfloor {\frac{n}{p}} \right\rfloor} \right)} \right]} \\=  &\frac{1}{2} (-1)^n {\left\lfloor {\frac{n}{p}} \right\rfloor} – \frac{1}{2} \begin{cases} {-\left\lfloor {\frac{n}{p}} \right\rfloor} &\textrm{if } p\equiv 0 \pmod{2} \\   {\left\lfloor {\frac{n}{p}} \right\rfloor} \bmod{2} &\textrm{if } p\equiv 1 \pmod{2} \end{cases} \quad p \in \mathbb{N} \end{align}

\begin{align} \sum\limits_{k=0}^{n}{k\left\lfloor {\frac{k}{p}} \right\rfloor} &=  \frac{1}{2}{\left\lfloor {\frac{n}{p}} \right\rfloor} \left[ {{n\,(n+1)}\, – \frac{p}{6} {\left({{\left\lfloor {\frac{n}{p}} \right\rfloor} + 1} \right)}  {\left( {{2\, p\left\lfloor {\frac{n}{p}} \right\rfloor} + p\,\, – 3}\right)}}\right]  \\ &=  \frac{{\left\lfloor {\frac{n}{p}} \right\rfloor} \left[ {{6\,n\,(n+1)}\, – {2\,p^2}{\left\lfloor {\frac{n}{p}} \right\rfloor}^2 } – 3\,p\,(p-1){\left\lfloor {\frac{n}{p}} \right\rfloor} – p\,(p-3)\right]}{12} , \quad p \in \mathbb{N} \end{align}

\begin{align} \sum\limits_{k=0}^{n}{(-1)^k k\left\lfloor {\frac{k}{p}} \right\rfloor} &= {\left\lfloor {\frac{n+1}{2}} \right\rfloor} {\left\lfloor {\frac{n}{p}} \right\rfloor} (-1)^n  +  \sum\limits_{j=1}^{\left\lfloor {\frac{n}{p}} \right\rfloor}{(-1)^{pj} \left\lfloor {\frac{pj}{2}} \right\rfloor} \\ \notag &=  \frac{2n+1}{4} {\left\lfloor {\frac{n}{p}} \right\rfloor}(-1)^n + \\ \notag &\quad + \begin{cases} \frac{p}{2} \frac{{\left\lfloor {\frac{n}{p}} \right\rfloor}\left(\left\lfloor {\frac{n}{p}} \right\rfloor + 1\right)}{2} – \frac{1}{4}{\left\lfloor {\frac{n}{p}} \right\rfloor}   &\textrm{if } p\equiv 0 \pmod{2} \\   \frac{p}{2} {\left\lfloor \frac{{\left\lfloor {\frac{n}{p}} \right\rfloor} + 1}{2} \right\rfloor} (-1)^{\left\lfloor {\frac{n}{p}} \right\rfloor} + \frac{1}{4} \frac{1 – (-1)^{\left\lfloor {\frac{n}{p}} \right\rfloor}}{2}   &\textrm{if } p\equiv 1 \pmod{2} \end{cases} \quad p \in \mathbb{N} \end{align}

\begin{equation} {S_{n}}^{(r)}:=\sum\limits_{k=1}^{n}{\left\lfloor {k^{r}} \right\rfloor} , \, r \in \mathbb{R}, \quad  r\gt{0} \end{equation}

\begin{align} \sum\limits_{k=1}^{n}{\left\lfloor {\sqrt{k}} \right\rfloor} &= \left( {n+1} \right) \left\lfloor {\sqrt{n}}\, \right\rfloor – S_{\left\lfloor {\sqrt{n}}\, \right\rfloor}^{(2)} \\ \notag &= \left( {n+1} \right) \left\lfloor {\sqrt{n}}\, \right\rfloor – \frac{{\left\lfloor {\sqrt{n}} \right\rfloor \left( {\left\lfloor {\sqrt{n}} \right\rfloor+1} \right) \left( {2\left\lfloor {\sqrt{n}} \right\rfloor+1} \right)}}{6}\end{align}

\begin{align} \sum\limits_{k=1}^{n}{\left\lfloor {\sqrt{k}} \right\rfloor}^2 &= \left( {n+1} \right) {\left\lfloor {\sqrt{n}}\, \right\rfloor}^2 + S_{\left\lfloor {\sqrt{n}}\, \right\rfloor}^{(2)} – 2\cdot S_{\left\lfloor {\sqrt{n}}\, \right\rfloor}^{(3)} \\ \notag &= \left( {n+1} \right) {\left\lfloor {\sqrt{n}}\, \right\rfloor}^2 + \frac{{\left\lfloor {\sqrt{n}} \right\rfloor \left( {\left\lfloor {\sqrt{n}} \right\rfloor+1} \right) \left( {2\left\lfloor {\sqrt{n}} \right\rfloor+1} \right)}}{6} \\ \notag &\quad – \frac{{{\left\lfloor {\sqrt{n}} \right\rfloor}^2 {\left( {\left\lfloor {\sqrt{n}} \right\rfloor+1} \right)}}^2}{2}  \end{align}

\begin{equation} \sum\limits_{k=1}^{n}{\left \lfloor {k^\frac{1}{3}} \right\rfloor} = \left( {n+1} \right) {\left \lfloor {n^\frac{1}{3}} \right\rfloor} -\frac{{\left \lfloor {n^\frac{1}{3}} \right\rfloor}^{2}\,{\left( {{\left \lfloor {n^\frac{1}{3}} \right\rfloor} +1} \right)}^2}{4}\end{equation}

\begin{equation} \sum\limits_{k=1}^{n}{\left \lfloor {k^\frac{1}{p}} \right\rfloor}=\left( {n+1} \right)\cdot {\left\lfloor {n^\frac{1}{p}} \right\rfloor} – \sum\limits_{k=1}^{\left\lfloor {n^\frac{1}{p}} \right\rfloor}{{k}^{p}}, \quad p\in\mathbb{R},\,p \ge {1}\end{equation}

\begin{equation} S_{n}^{\left(\frac{1}{p}\right)}+{S}_{\left\lfloor {n^\frac{1}{p}} \right\rfloor}^{(p)} = \left( {n+1} \right)\cdot \left \lfloor {n^\frac{1}{p}} \right\rfloor, \quad p\in\mathbb{R},\, p \ge {1} \end{equation}

\begin{align} \sum\limits_{k=1}^{n}{k\left\lfloor {{k^{\frac{1}{p}}}} \right\rfloor} = \frac{1}{2} \left( { n(n+1) {\left\lfloor {n^{\frac{1}{p}}} \right\rfloor} + S_{\left\lfloor {n^{\frac{1}{p}}} \right\rfloor}^{(p)} – S_{\left\lfloor {n^{\frac{1}{p}}}\right\rfloor}^{(2p)} }\right) \end{align}

\begin{align} \sum\limits_{k=1}^{n}{(-1)^k\,\left \lfloor {k^\frac{1}{p}} \right\rfloor} &= \frac{1}{4} \left( 2 \,(-1)^n {\left\lfloor {n^\frac{1}{p}} \right\rfloor} + {(-1)^{\left\lfloor {n^\frac{1}{p}} \right\rfloor} – 1} \right) \\ \notag &= \frac{1}{2} \left( (-1)^n {\left\lfloor {n^\frac{1}{p}} \right\rfloor} + {(-1)^{\left\lfloor {n^\frac{1}{p}} \right\rfloor}}\bmod {2} \right), \quad p\in\mathbb{R},\, p \ge {1}  \end{align}

\begin{equation}\sum\limits_{k=1}^{n}{\left\lfloor {\log_2{(k)}} \right\rfloor} = ({n+1}) \left\lfloor {\log_2{(n)}} \right\rfloor – 2 \left({2^{ \left\lfloor {\log_2{(n)}} \right\rfloor} -1} \right)\end{equation}

\begin{equation}\sum\limits_{k=1}^{n}{\left\lfloor {\log{(k)}} \right\rfloor} = ({n+1}) \left\lfloor {\log{(n)}} \right\rfloor – \frac{10}{9} \left({10^{ \left\lfloor {\log{(n)}} \right\rfloor} -1} \right)\end{equation}

\begin{align}\sum\limits_{{k=1}}^{n}{\left\lfloor {\log_p{(k)}} \right\rfloor} &= ({n+1}) \left\lfloor {\log_p{(n)}} \right\rfloor – \frac{p}{p-1} \left({p^{ \left\lfloor {\log_p{(n)}} \right\rfloor} -1} \right) \\ \notag &\quad  p\in \mathbb{N},  \, p \gt 1 \end{align}

\begin{align}\sum\limits_{{k=1}}^{n}{k\,\left\lfloor {\log_p{(k)}} \right\rfloor} &= \frac{n(n+1)}{2} \left\lfloor {\log_p{(n)}} \right\rfloor – \frac{p}{2} \frac{\left({p^{ \left\lfloor {\log_p{(n)}} \right\rfloor} -1} \right) \left({p^{ \left\lfloor {\log_p{(n)}} \right\rfloor + 1} -1} \right)}{p^2-1} \\ \notag &\quad  p\in \mathbb{N},  \, p \gt 1 \end{align}

\begin{equation}\sum\limits_{{k=1}}^{n}{(-1)^k\,\left\lfloor {\log_p{(k)}} \right\rfloor} = \frac{(-1)^p+(-1)^n}{2} \left\lfloor {\log_p{(n)}} \right\rfloor,  \quad  p\in \mathbb{N},  \, p \gt 1 \end{equation}

   

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