Given-that-S-n-a-1-r-n-1-r-r-1-show-that-S-3n-S-2n-S-n-r-2n-hence-given-that-r-1-2-find-n-0-S-3n-S-2n-S-n-

Question Number 66107 by Rio Michael last updated on 09/Aug/19

G i v e n t h a t S_{n} = \frac{a (1 - r^{n})}{1 - r}, r \neq 1, s h o w t h a t \frac{S_{3 n} - S_{2 n}}{S_{n}} = r^{2 n}

h e n c e g i v e n t h a t r = \frac{1}{2} f i n d \underset{n = 0}{\sum^{\infty}} (\frac{S_{3 n} - S_{2 n}}{S_{n}})

Commented by Prithwish sen last updated on 09/Aug/19

\frac{S_{3 n} - S_{2 n}}{S_{n}} = \frac{1 - r^{3 n} - 1 + r^{2 n}}{1 - r^{n}} = \frac{r^{2 n} (1 - r^{n})}{(1 - r^{n})} = r^{2 n}

\underset{n = 0}{\sum^{\infty}} r^{2 n} = {(\frac{1}{2})}^{0} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{4} + \dots . . = \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}

please check .

Commented by Rio Michael last updated on 09/Aug/19

c o r r e c t s i r t h a n k s

Commented by mathmax by abdo last updated on 10/Aug/19

w e h a v e \frac{S_{3 n} - S_{2 n}}{S_{n}} = \frac{a (1 - r^{3 n}) - a (1 - r^{2 n})}{a (1 - r^{n})} = \frac{r^{2 n} - r^{3 n}}{1 - r^{n}}

= \frac{r^{2 n} (1 - r^{n})}{1 - r^{n}} = r^{2 n} w i t h r \neq 1

\Rightarrow \sum_{n = 0}^{\infty} \frac{S_{3 n} - S_{2 n}}{S_{n}} = \sum_{n = 0}^{\infty} r^{2 n} = \frac{1 - r^{2 (n + 1)}}{1 - r^{2}}

a n d f o r ∣ r ∣< 1 w e g e t \sum_{n = 0}^{\infty} r^{2 n} = \frac{1}{1 - r^{2}} \Rightarrow

\sum_{n = 0}^{\infty} \frac{S_{3 n} - S_{2 n}}{S_{n}} = \frac{1}{1 - (\frac{1}{4})} = \frac{4}{3}