TITLE: Closed form for the sum $\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$
QUESTION [9 upvotes]: I want a closed form for the sum $$S=\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$$
where: $a\ne 1<p<n;\quad p\in\mathbb Z$
I know a related identity, 
$$\quad\displaystyle \sum_{k=0}^n\left\lfloor\frac{k}{p}\right\rfloor=\left(n+1-\frac{p}{2}\right)\left\lfloor\frac{n}{p}\right\rfloor-\frac{p}{2}\left\lfloor\frac{n}{p}\right\rfloor^2\quad\;,$$ 
but I can't apply it to this problem!
Result for this sum: 
$$\quad\displaystyle S=\dfrac{a^{n+1}}{a-1}\left\lfloor\frac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}$$
Can you prove it?

REPLY [3 votes]: My Solution
Apply the Summation by parts, we have:
\begin{align*} S=\sum_{k=1}^n a^k\left\lfloor\dfrac{k}{p}\right\rfloor&=\dfrac{1}{a-1}\sum_{k=1}^n\left\lfloor\dfrac{k}{p}\right\rfloor\Delta(a^k)
\\ &= \left.\dfrac{1}{a-1}\cdot a^k\left\lfloor\dfrac{k}{p}\right\rfloor\right|_{k=1}^{n+1}-\dfrac{1}{a-1}\sum_{k=1}^n a^{k+1}\left(\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor\right) \\ &= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{k=1}^n a^{k+1}\left(\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor\right)\end{align*}
Note that: $\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor=\begin{cases}1,\quad p\mid k+1\\0,\quad p\nmid k+1\end{cases}=\begin{cases}1,\quad k=mp-1\\0,\quad otherwise\end{cases}$
Therefore:
\begin{align*}S&=\dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{1\le mp-1\le n} a^{mp}\\ &=\dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor} \dfrac{a^{p(m+1)}-a^{pm}}{a^p-1}
\\&= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{(a-1)(a^p-1)}\sum_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor} \Delta[a^{pm}]
\\ &= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\left.\dfrac{a^{pm}}{(a-1)(a^p-1)}\right|_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor+1}\\&= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}\end{align*}