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How can it be demonstrate that p^n = p^n-1 + p^n-2 ... + p + 1?

Anonymous Sat 05 Feb 17:25:15 2022 No.14190293 View Reply Original Report

Quoted By:

Anonymous

Anonymous Sat 05 Feb 2022 17:29:07 No.14190306 Report

Quoted By: >>14190328 >>14192194

$\displaystyle \boxed{0 p^n-1 = (p-1)(p^{n-1}+p^{n-2}+ \dots +p+1) \\ \dfrac{p^n-1}{p-1} = \sum \limits_{j=0}^{n-1}p^j \\ \displaystyle \lim_{n \to \infty} \dfrac{p^n-1}{p-1} = \lim_{n \to \infty} \sum \limits_{j=0}^{n-1}p^j \\ \displaystyle \dfrac{0-1}{p-1} = \sum \limits_{j=0}^{\infty}p^j \implies \dfrac{1}{1-p} = \sum \limits_{j=0}^{\infty}p^j $

Anonymous

Anonymous Sat 05 Feb 2022 17:35:08 No.14190328 Report

Quoted By:

>>14190306
Sorry, I meant to write ">", not equal

Anonymous

Anonymous Sun 06 Feb 2022 00:24:09 No.14191991 Report

Quoted By:

What is given?
p? or n?
either way you may use mathematical induction

Anonymous

Anonymous Sun 06 Feb 2022 01:23:20 No.14192194 Report

Quoted By: >>14192250

>>14190306
$p^n = p^{n-1} ... + p + 1$
$p^n - p^{n-1} ... - p = 1$
p is either prime or divisible by numbers $a,b$
If p is prime, then $p^n-p^{n-1}...-p$ is divisible by p, yielding $p^{n-1}-p^{n-2}...- p - 1$ .
Therefore 1 must be divisible by $p$ which is not possible.
The same proof applies to $a,b$ therefore p can't be a natural number.
If p is a fractional number, set $p = \frac{a}{b}$
$\frac{a^n}{b^n} = \frac{a^{n-1}}{b^{n-1}}+ ... + \frac{a}{b} + 1$ . Multiply by $b^n$ .
$a^n = a^{n-1}b+ ... + ab^{n-1} + b^n$
$a^n - a^{n-1}b- ... - ab^{n-1} = b^n$
Suppose that $a,b$ are relatively prime to each other.
Then dividing the left side by $a$ yields the desired contradiction.
If $a,b$ are not relatively prime, divide all the shared factors yielding the relative primes $c,d \rightarrow$
$c^n = c^{n-1}d+ ... + cd^{n-1} + d^n$ .
Therefore, p can't be a rational number either. p is thus can only be an irrational number for any n.

Anonymous

Anonymous Sun 06 Feb 2022 01:34:49 No.14192250 Report

Quoted By:

>>14192194
We know that p can't be rational. More generally

$\frac{p^n-1}{p-1} = \sum_{j=0}^{n-1}{p^j}$
$\frac{p^n-1}{p-1} = p^n$
$p^n-1 = p^{n+1}-p^n$
$2p^n = p^{n+1}+1$

The solution of $2p^n = p^{n+1}+1$ yields the solution for $p^n = p^{n-1}+p^{n-2}+p+1$

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