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

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 Σ_(n=0) ^∞ (((S_(3n) −S_(2n) )/S_n ))

$${Given}\:{that}\:{S}_{{n}} \:=\:\frac{{a}\left(\mathrm{1}\:−{r}^{{n}} \right)}{\mathrm{1}−{r}}\:,\:{r}\:\neq\:\mathrm{1},\:{show}\:{that}\:\frac{{S}_{\mathrm{3}{n}} \:−{S}_{\mathrm{2}{n}} }{{S}_{{n}} \:}\:=\:{r}^{\mathrm{2}{n}} \\ $$$${hence}\:{given}\:{that}\:{r}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{find}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{S}_{\mathrm{3}{n}} \:−{S}_{\mathrm{2}{n}} }{{S}_{{n}} }\right) \\ $$

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

((S_(3n) −S_(2n) )/S_n ) = ((1−r^(3n) −1+r^(2n) )/(1−r^n )) = ((r^(2n) (1−r^n ))/((1−r^n ))) = r^(2n) Σ_(n=0) ^∞ r^(2n) =((1/2))^0 +((1/2))^2 +((1/2))^4 +..... = (1/(1−(1/4))) = (4/3) please check.

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

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

correct sir thanks

$${correct}\:{sir}\:{thanks} \\ $$

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

we have ((S_(3n) −S_(2n) )/S_n ) =((a(1−r^(3n) )−a(1−r^(2n) ))/(a(1−r^n ))) =((r^(2n) −r^(3n) )/(1−r^n )) =((r^(2n) (1−r^n ))/(1−r^n )) =r^(2n) with r≠1 ⇒Σ_(n=0) ^∞ ((S_(3n) −S_(2n) )/S_n ) =Σ_(n=0) ^∞ r^(2n) =((1−r^(2(n+1)) )/(1−r^2 )) and for ∣r∣<1 we get Σ_(n=0) ^∞ r^(2n) =(1/(1−r^2 )) ⇒ Σ_(n=0) ^∞ ((S_(3n) −S_(2n) )/S_n ) =(1/(1−((1/4)))) =(4/3)

$${we}\:{have}\:\frac{{S}_{\mathrm{3}{n}} −{S}_{\mathrm{2}{n}} }{{S}_{{n}} }\:=\frac{{a}\left(\mathrm{1}−{r}^{\mathrm{3}{n}} \right)−{a}\left(\mathrm{1}−{r}^{\mathrm{2}{n}} \right)}{{a}\left(\mathrm{1}−{r}^{{n}} \right)}\:=\frac{{r}^{\mathrm{2}{n}} −{r}^{\mathrm{3}{n}} }{\mathrm{1}−{r}^{{n}} } \\ $$$$=\frac{{r}^{\mathrm{2}{n}} \left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}^{{n}} }\:={r}^{\mathrm{2}{n}} \:\:\:{with}\:{r}\neq\mathrm{1} \\ $$$$\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{S}_{\mathrm{3}{n}} −{S}_{\mathrm{2}{n}} }{{S}_{{n}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{r}^{\mathrm{2}{n}} \:=\frac{\mathrm{1}−{r}^{\mathrm{2}\left({n}+\mathrm{1}\right)} }{\mathrm{1}−{r}^{\mathrm{2}} } \\ $$$${and}\:{for}\:\mid{r}\mid<\mathrm{1}\:\:\:\:{we}\:{get}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{r}^{\mathrm{2}{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{r}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{S}_{\mathrm{3}{n}} −{S}_{\mathrm{2}{n}} }{{S}_{{n}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\frac{\mathrm{4}}{\mathrm{3}} \\ $$

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