Consecutive sum of exponentiations

[scientifico]

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

Thank you!

 

[JJacquelin]

Simply expand :
(x^0+x^1+x^2+...+x^n)(x-1)

 

[scientifico]

I didn't understand why you wrote (x-1) after the expansion... thank you

 

[HallsofIvy]

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 x^0+ x^1+ x^2+ \cdot\cdot\cdot+ x^n is a geometric series with "common factor" x. The sum of a finite geometric series is
\frac{1- x^{n+1}}{1- x}= \frac{x^{n+1}- 1}{x- 1}.

 

[JJacquelin]

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 !