Talk:Sum of Geometric Sequence

I'm partial to using sum notation, since unnecessary indefinite expansions with $\cdots$ seems somehow less exact since it's easier to lose track of something. I'm happy to defer to what others think, mostly I'm curious what the accepted custom is. I wrote out the proof with sums now while I was thinking of it.

Let $S_n = \sum_{j = 0}^{n - 1} x^j$.

Then $x S_n = x \sum_{j = 0}^{n - 1} x^j = \sum_{j = 0}^{n - 1} x\cdot x^j = \sum_{j = 1}^n x^j$.

Then $S_n(x - 1) = x S_n - S_n = \sum_{j = 1}^{n} x^j -\sum_{j = 0}^{n - 1} x^j = x^n - x^0 = x^n - 1$.

Thus $\sum_{j = 0}^{n - 1} x^j = S_n = \frac{x^n - 1}{x - 1}$ for $x \neq 1$.

--Cynic 03:32, 25 November 2008 (UTC)

Certainly, feel free to add it. I was in two minds about this one, the $\cdots$ style is elementary and simple to follow and may be accessed by students at an elementary level and is arguably more accessible. But no problem having both versions in. Thoughts? --Matt Westwood 06:23, 25 November 2008 (UTC)

... BTW I might make it go:

$S_n(x - 1) = x S_n - S_n = \sum_{j = 1}^{n} x^j -\sum_{j = 0}^{n - 1} x^j = x^n + \sum_{j = 1}^{n-1} x^j - \left({x^0 + \sum_{j = 1}^{n - 1} x^j}\right) = x^n - x^0 = x^n - 1$

to make it blindingly obvious ... --Matt Westwood 06:27, 25 November 2008 (UTC)