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

$$Giventhat{S}_n=\frac{a(1-r^n)}{1-r},r\ne 1,showthat\frac{S_{3n}-S_{2n}}{S_n}=r^{2n}$$$$hencegiventhatr=\frac{1}{2}find\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\frac{S_{3n}-S_{2n}}{S_n}\right)$$

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

$$\frac{{\mathrm{S}}_{3\mathrm{n}}-{\mathrm{S}}_{2\mathrm{n}}}{{\mathrm{S}}_{\mathrm{n}}}=\frac{1-{\mathrm{r}}^{3\mathrm{n}}-1+{\mathrm{r}}^{2\mathrm{n}}}{1-{\mathrm{r}}^{\mathrm{n}}}=\frac{{\mathrm{r}}^{2\mathrm{n}}(1-{\mathrm{r}}^{\mathrm{n}})}{(1-{\mathrm{r}}^{\mathrm{n}})}={\mathrm{r}}^{2\mathrm{n}}$$$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{r}}^{2\mathrm{n}}={\left(\frac{1}{2}\right)}^0+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^4+.....=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$$$$\mathrm{please}\mathrm{check}.$$

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

$$correctsirthanks$$

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

$$wehave\frac{S_{3n}-S_{2n}}{S_n}=\frac{a(1-r^{3n})-a(1-r^{2n})}{a(1-r^n)}=\frac{r^{2n}-r^{3n}}{1-r^n}$$$$=\frac{r^{2n}(1-r^n)}{1-r^n}=r^{2n}withr\ne 1$$$$\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{S_{3n}-S_{2n}}{S_n}=\sum _{n=0}^{\mathrm{\infty }}{r}^{2n}=\frac{1-r^{2(n+1)}}{1-r^2}$$$$andfor\mid r\mid <1weget\sum _{n=0}^{\mathrm{\infty }}{r}^{2n}=\frac{1}{1-r^2}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{S_{3n}-S_{2n}}{S_n}=\frac{1}{1-\left(\frac{1}{4}\right)}=\frac{4}{3}$$

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