sequences $b^{2n}-1$ and $b^{2n-1}+1$ are divisible by $b+1$

Consider the alternating geometric finite series

$S_{m+1}(\mu )={\displaystyle \sum _{i=0}^m}{(-1)}^{i+\mu }{b}^i,$ (1)

where $\mu =1,2$ and $b\ge 2$ an integer. Multiplying (1) by $-b$ and subtracting from it

$(b+1)S_{m+1}(\mu )={\displaystyle \sum _{i=0}^m}{(-1)}^{i+\mu }{b}^i-{\displaystyle \sum _{i=0}^m}{(-1)}^{i+1+\mu }{b}^{i+1}$

and by elemental manipulations, we obtain

$S_{m+1}(\mu )={\displaystyle \frac{{(-1)}^{\mu }[1-{(-1)}^{m+1}{b}^{m+1}]}{b+1}}={\displaystyle \sum _{i=0}^m}{(-1)}^{i+\mu }{b}^i.$ (2)

Let $\mu =1$, $m=2n-1$. Then

${\displaystyle \frac{b^{2n}-1}{b+1}}=-{\displaystyle \sum _{i=0}^{2n-1}}{(-1)}^i{b}^i.$ (3)

Likewise, for $\mu =2$, $m=2n-2$

${\displaystyle \frac{b^{2n-1}+1}{b+1}}={\displaystyle \sum _{i=0}^{2n-2}}{(-1)}^i{b}^i,$ (4)

as desired.

Palindromic numbers of even length

As an application of above sequences, let us consider an even palindromic number (EPN) of arbitrary length $2n$ which can be expressed in any base $b$ as

${(EPN)}_n={\displaystyle \sum _{k=0}^{n-1}}{b}_k(b^{2n-1-k}+b^k)={\displaystyle \sum _{k=0}^{n-1}}{\displaystyle \frac{b_k}{b^k}}[(b^{2n-1}+1)+(b^{2k}-1)],$ (5)

where $0\le b_k\le b-1$.It is clear, from (3) and (4), that ${(EPN)}_n$ is divisible by $b+1$. Indeed this one can be given by

${(EPN)}_n=(b+1){\displaystyle \sum _{k=0}^{n-1}}{\displaystyle \sum _{j=0}^{2(n-1-k)}}{b}_k{(-1)}^j{b}^{k+j}.$ (6)

Title	sequences $b^{2n}-1$ and $b^{2n-1}+1$ are divisible by $b+1$
Canonical name	SequencesB2n1AndB2n11AreDivisibleByB1
Date of creation	2013-03-22 16:14:19
Last modified on	2013-03-22 16:14:19
Owner	perucho (2192)
Last modified by	perucho (2192)
Numerical id	6
Author	perucho (2192)
Entry type	Derivation
Classification	msc 11A63