Power series $ \sum_{r=1}^{n}x^{r}=\:?$

Question

I want to know a formula for

$$\displaystyle \sum_{r=1}^{n}x^{r}=\:?$$

I can't say i can see where to derive it from at all. Any help or pointers would be greatly appreciated. Thanks

Hint: Geometric progression. – njguliyev Oct 20 '13 at 19:44 — njguliyev, Oct 20 '13 at 19:44
Almost surely this has been answered here before – Jean-Sébastien Oct 20 '13 at 19:52 — Jean-Sébastien, Oct 20 '13 at 19:52

score 1 · Answer 1 · edited Oct 20 '13 at 19:53

1

Hint: Write

$$S = \sum_{i=1}^{n} x^{i} = 1 + x + x^{2} + \cdots x^{n}$$

$$Sx = x + x^{2} + \cdots + x^{n+1}$$

So $$S - Sx= 1 - x^{n+1}$$

Can you try and solve for $S$ from here?

edited Oct 20 '13 at 19:53

Pedro

122,002

answered Oct 20 '13 at 19:46

Elchanan Solomon

30,807

Oh, very nice! That's just what i wanted to see. Man i feel stupid. It's been a long day. Thanks for making it so clear and simple. – AntonySC Oct 20 '13 at 20:55

score 1 · Answer 2 · answered Oct 20 '13 at 19:55

here is a way to do this we have

$$ (1-x)(x + x^2 + x^3 + \cdots + x^n) = x + x^2 + x^3 + \cdots + x^n - x(x + x^2 + x^3 + \cdots + x^n) $$

$$ = x + x^2 + x^3 + \cdots + x^n - (x^2 + x^3 + \cdots x^{n+1}) = x -x^{n+1}$$

$$ \Rightarrow x + x^2 + x^3 + \cdots + x^n = \frac{x^{n+1} - x}{x-1} $$

score 0 · Answer 3 · answered Oct 20 '13 at 19:47

0

Let $$ S=\sum_{r=1}^{n}x^r. $$ Note that $$ xS=\sum_{r=1}^{n}x^{r+1}=\sum_{r=2}^{n+1}x^r, $$ so that $$ (1-x)S=\sum_{r=1}^{n}x^r-\sum_{r=2}^{n+1}x^r=1-x^{n+1}. $$ So, $$ S=\frac{1-x^{n+1}}{1-x}. $$

answered Oct 20 '13 at 19:47

Nick Peterson

32,430

Power series $ \sum_{r=1}^{n}x^{r}=\:?$

3 Answers3