If the sum of the series displaystyle∑n=0∞ rn,|r|

Question

If the sum of the series displaystyle∑n=0∞ rn,|r|<1, is s, then sum of the series displaystyle∑n=0∞ r2 n is (A) s2 (B) (2 s/s2-1) (C) (s2/2 s+1) (D) (s2/2 s-1)

Tardigrade Admin · Accepted Answer

s = displaystyle∑n=0∞ rn=1+r+r2+r3+⋅s text to ∞ =(1/1-r) ⇒ r=1-(1/s)=(s-1/s) displaystyle∑n=0∞ r2 n=(1/1-r2)=(1/1-((s-1)2)s2)=(s2/2 s-1)

Q. If the sum of the series $n = 0 \sum \infty r^{n}, ∣ r ∣ < 1,$ is $s,$ then sum of the series $n = 0 \sum \infty r^{2 n}$ is

$s^{2}$

$\frac{2 s}{s ^{2} - 1}$

$\frac{s ^{2}}{2 s + 1}$

$\frac{s ^{2}}{2 s - 1}$

$s = n = 0 \sum \infty r^{n} = 1 + r + r^{2} + r^{3} + \dots to \infty$
$= \frac{1}{1 - r}$
$\Rightarrow r = 1 - \frac{1}{s} = \frac{s - 1}{s}$
$n = 0 \sum \infty r^{2 n} = \frac{1}{1 - r ^{2}} = \frac{1}{1 - \frac{( s - 1 ) ^{2}}{s ^{2}}} = \frac{s ^{2}}{2 s - 1}$

Q. If the sum of the series n=0∑∞​rn,∣r∣<1, is s, then sum of the series n=0∑∞​r2n is

s2

s2−12s​

2s+1s2​

2s−1s2​