# 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

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

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

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