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

Consider the alternating geometric finite series

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

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

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

and by elemental manipulations, we obtain

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

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

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

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

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

 $\displaystyle(EPN)_{n}=\sum_{k=0}^{n-1}b_{k}(b^{2n-1-k}+b^{k})=\sum_{k=0}^{n-1% }\frac{b_{k}}{b^{k}}[(b^{2n-1}+1)+(b^{2k}-1)],$ (5)

where $0\leq b_{k}\leq b-1$.It is clear, from (3) and (4), that $(EPN)_{n}$ is divisible by $b+1$. Indeed this one can be given by

 $\displaystyle(EPN)_{n}=(b+1)\sum_{k=0}^{n-1}\sum_{j=0}^{2(n-1-k)}b_{k}(-1)^{j}% b^{k+j}.$ (6)
Title sequences $b^{2n}-1$ and $b^{2n-1}+1$ are divisible by $b+1$ SequencesB2n1AndB2n11AreDivisibleByB1 2013-03-22 16:14:19 2013-03-22 16:14:19 perucho (2192) perucho (2192) 6 perucho (2192) Derivation msc 11A63