Consecutive sum of exponentiations

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Discussion Overview

The discussion revolves around the formula for the sum of exponentiations, specifically the expression x^0 + x^1 + x^2 + ... + x^n, and how to demonstrate the formula \(\frac{x^{n+1}-1}{x-1}\). Participants explore different approaches to understanding and proving this formula.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant asks for a demonstration of the formula for the sum of exponentiations.
  • Another participant suggests expanding the expression (x^0 + x^1 + x^2 + ... + x^n)(x - 1) as a method to understand the formula.
  • A different participant questions the relevance of multiplying by (x - 1) in the context of the expansion.
  • One participant explains that the multiplication by (x - 1) relates to the denominator in the original formula and provides reasoning based on the properties of geometric series.
  • Another participant reiterates the suggestion to multiply the series by (x - 1) to simplify and understand the derivation of the formula.

Areas of Agreement / Disagreement

Participants express differing views on the approach to demonstrating the formula, with some supporting the multiplication method and others questioning its necessity. The discussion remains unresolved regarding the best method to demonstrate the formula.

Contextual Notes

Participants have not reached a consensus on the most effective method for demonstrating the formula, and there are varying interpretations of the steps involved in the expansion and simplification process.

scientifico
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Hello, i read that a sum of exponentiations like [itex]x^0+x^1+x^2+x^3...+x^n[/itex] can be solved with this forumula [itex]\frac{x^(n+1)-1}{x-1}[/itex], how is it possible do demonstrate this resolutive formula?

Thank you!
 
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Simply expand :
(x^0+x^1+x^2+...+x^n)(x-1)
 
I didn't understand why you wrote (x-1) after the expansion... thank you
 
Because of the x- 1 in the denominator in your equation.
If A= B/(x- 1) then A(x- 1)= B.

Is that true here?
x
That is not, however, how I would handle this problem. I would note that [itex]x^0+ x^1+ x^2+ \cdot\cdot\cdot+ x^n[/itex] is a geometric series with "common factor" x. The sum of a finite geometric series is
[tex]\frac{1- x^{n+1}}{1- x}= \frac{x^{n+1}- 1}{x- 1}[/tex].
 
scientifico said:
I didn't understand why you wrote (x-1) after the expansion... thank you

You will understand if you make the multiplication of the series by (x-1) and then simplify. Just do it !
 

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