La forma cerrada para la suma de $\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$

Question

Quiero una forma cerrada para la suma de $$S=\sum_{k=0}^n a^k \left\lfloor\frac{k}{p}\right\rfloor$$ donde: $a\ne 1<p<n;\quad p\in\mathbb Z$

Sé relacionados con la identidad,

$$\quad\displaystyle \sum_{k=0}^n\left\lfloor\frac{k}{p}\right\rfloor=\left(n+1-\frac{p}{2}\right)\left\lfloor\frac{n}{p}\right\rfloor-\frac{p}{2}\left\lfloor\frac{n}{p}\right\rfloor^2\quad\;,$$

pero no puedo aplicarlo a este problema!

El resultado de esta suma:

$$\quad\displaystyle S=\dfrac{a^{n+1}}{a-1}\left\lfloor\frac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}$$

Puedes probarlo?

Answer 1

Escribir $n=cp+r$,$c,r\in\mathbb N\cup\{0\}$, e $r<p$. Entonces, para $k\leq n$, $$\left\lfloor\frac kp\right\rfloor=\begin{cases}0,&\ 0\leq k<p\\ 1,&\ p\leq k<2p,\\ \vdots\\ m,&\ mp\leq k<(m+1)p\\ \vdots\\ c,&\ cp\leq k\leq n \end{casos}$$ Entonces $$S=\sum_{k=0}^na^k\,\left\lfloor\frac kp\right\rfloor=\sum_{m=0}^{c-1}\sum_{k=mp}^{(m+1)p-1}ma^k\,+\sum_{k=cp}^{n}ca^k =\sum_{m=1}^{c-1}m\,\sum_{k=mp}^{(m+1)p-1}a^k\,+\sum_{k=cp}^{n}ca^k\\ =\sum_{m=1}^{c-1}m\,\frac{a^{mp}-a^{(m+1)p}}{1}+c\,\frac{a^{cp}-a^{n+1}}{1}.$$ Tenga en cuenta que en la primera suma, si juntamos el término negativo para el índice de $m$ y el positivo para $m+1$, obtenemos $$-ma^{(m+1)p}+(m+1)^{(m+1)p}=a^{(m+1)p}.$$ Así $$(1-a)S=\sum_{m=1}^ca^{mp}-ca^{n+1}=\frac{a^p^{(c+1)p}}{1-a^p}-ca^{n+1},$$ o $$S=\frac{ca^{n+1}}{a-1}-\frac{a^{(c+1)p}-a^p}{(a^p-1)(a-1)}=\frac{ca^{n+1}}{a-1}-\frac{a^p(a^{pc}-1)}{(a^p-1)(a-1)}.$$ Donde $$c=\left\lfloor \frac np\right\rfloor.$$

Answer 2

Mi Solución

Aplicar la Suma por partes, tenemos: $$\begin{align*} S=\sum_{k=1}^n a^k\left\lfloor\dfrac{k}{p}\right\rfloor&=\dfrac{1}{a-1}\sum_{k=1}^n\left\lfloor\dfrac{k}{p}\right\rfloor\Delta(a^k) \\ &= \left.\dfrac{1}{a-1}\cdot a^k\left\lfloor\dfrac{k}{p}\right\rfloor\right|_{k=1}^{n+1}-\dfrac{1}{a-1}\sum_{k=1}^n a^{k+1}\left(\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor\right) \\ &= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{k=1}^n a^{k+1}\left(\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor\right)\end{align*}$$

Tenga en cuenta que: $\left\lfloor\dfrac{k+1}{p}\right\rfloor-\left\lfloor\dfrac{k}{p}\right\rfloor=\begin{cases}1,\quad p\mid k+1\\0,\quad p\nmid k+1\end{cases}=\begin{cases}1,\quad k=mp-1\\0,\quad otherwise\end{cases}$

Por lo tanto: $$\begin{align*}S&=\dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{1\le mp-1\le n} a^{mp}\\ &=\dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{a-1}\sum_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor} \dfrac{a^{p(m+1)}-a^{pm}}{a^p-1} \\&= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{1}{(a-1)(a^p-1)}\sum_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor} \Delta[a^{pm}] \\ &= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\left.\dfrac{a^{pm}}{(a-1)(a^p-1)}\right|_{m=1}^{\left\lfloor\frac{n+1}{p}\right\rfloor+1}\\&= \dfrac{a^{n+1}}{a-1}\left\lfloor\dfrac{n+1}{p}\right\rfloor-\dfrac{a^p\left(a^{p\left\lfloor\frac{n+1}{p}\right\rfloor}-1\right)}{(a-1)(a^p-1)}\end{align*}$$

Answer 3

Llame a la suma anterior $S_n$ para mayor claridad.

Al $n=pm-1$ $$S_{pm-1} = \sum_{l=0}^{m-1} \left(la^{lp}\sum_{i=0}^{p-1} a^i\right) = \frac{a^p-1}{a-1}\sum_{l=0}^{m-1}la^{lp}$$

Usted puede encontrar la forma cerrada para $\sum_{l=0}^{m-1}la^{lp}$.

Al $n=pm-1+j$ $j=0,1,p-1$ - que es$m=\left\lfloor\frac{n+1}{p}\right\rfloor$$j=n-pm+1$, entonces:

$$S_m = S_{pm-1} + m\sum_{k=0}^{j-1} a^{pm+k}=S_{pm-1}+ma^{pm}\frac{a^j-1}{a-1}$$

Así que finalmente sólo se necesita una forma cerrada para$\sum_{l=0}^{m-1}lz^l$, $z$ multiplicado por la derivada de la $\frac{z^{m}-1}{z-1}$. Luego sustituido $z=a^p$.