Question 1163948
<font face="Times New Roman" size="+2">


The *[tex \Large n]th partial sum of a geometric sequence is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a\(1\,-\,r^n\)}{\(1\,-\,r\)]


Then, given that *[tex \Large |r|\,<\,1], dividing the partial sum by the infinite sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sum_{i=0}^\infty\ ar^i\ =\ \frac{a}{1\,-\,r}]


yields


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ r^n]


Which is consistent with your calculations, that is you got *[tex \Large 1\,-\,r^2\ =\ \frac{5}{9}] from which you got *[tex \Large r\ =\ \pm\frac{2}{3}] using the partial sum *[tex \Large S_2].

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>