Question

The value of $\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\cdots \cdots \infty$ is

• A. $\frac{19}{24}$
• B. $\frac{11}{24}$
• C. $\frac{13}{24}$
• D. $\frac{23}{24}$

Answer 1

The correct option is A $\frac{19}{24}$

We have,
$\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\cdots \cdots \infty$

$=(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\cdots \infty )+(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots \cdots \infty )$ ---(1)

Which are in G.P.

lets take,$(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\cdots \infty )$

Here, $a=\frac{1}{2}$,---(2)

$r=\frac{\frac{1}{2^3}}{\frac{1}{2}}$

$r=\frac{1}{2^2}$ --(3)

lets take,

$(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots \cdots \infty )$

Here, $a=\frac{3}{2}$ --(4)

$r=\frac{\frac{1}{3^4}}{\frac{1}{3^2}}$

$r=\frac{1}{3^2}$ ---(5)

Now, lets take,

$(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\cdots \infty )+(\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots \cdots \infty )$

$=\left(\frac{\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2^2}\right)}\right)+\left(\frac{\left(\frac{1}{3^2}\right)}{1-\left(\frac{1}{3^2}\right)}\right)$

[Since, from (2),(3),(4) and (5)]

$=\frac{2}{3}+\frac{1}{8}$

$=\frac{19}{24}$

$\therefore \frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\cdots \cdots \infty =\frac{19}{24}$