Question

The value of $1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+\cdots \text{upto}\infty$ is

• A. $\frac{27}{14}$
• B. $\frac{21}{13}$
• C. $\frac{49}{27}$
• D. $\frac{256}{147}$

Answer 1

The correct option is C $\frac{49}{27}$

Let $S=1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+\cdots \text{upto}\infty ...\left(1\right)$
$\frac{1}{7}S=\frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^3}+\frac{16}{7^4}+\frac{25}{7^5}+\cdots \text{upto}\infty ...\left(2\right)$
Subtracting eqn $\left(2\right)$ from eqn $\left(1\right)$, we get
$\frac{6}{7}S=1+\frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+\cdots \text{upto}\infty ...\left(3\right)$
Multiplying eqn $\left(3\right)$ by $\frac{1}{7}$, we get
$\frac{6}{7^2}S=\frac{1}{7}+\frac{3}{7^2}+\frac{5}{7^3}+\frac{7}{7^4}+\cdots \text{upto}\infty ...\left(4\right)$
Subtracting eqn $\left(4\right)$ from eqn $\left(3\right)$, we get
$\frac{36}{49}S=1+\frac{2}{7}+\frac{2}{7^2}+\frac{2}{7^3}+\cdots \text{upto}\infty$
$\Rightarrow \frac{36}{49}S=1+\frac{2}{7}\left(\frac{1}{1-1/7}\right)$
$\therefore S=\frac{49}{27}$