Question

The infinite sum $1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+......$ equals

• 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 $x=1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+..$
Dividing both the sides by 7
$\frac{x}{7}=\frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^3}+\frac{16}{7^4}+...$
Subtracting iind from the 1st
$\frac{6x}{7}=1+\frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}...$
The RHS here is an arithmetic-Geometric series , Now divide the equation by 7 again
$\frac{6x}{49}=\frac{1}{7}+\frac{3}{7^2}+\frac{5}{7^3}+\frac{7}{7^4}$
Again subtracting if from the second one
$\Rightarrow \frac{36x}{49}-1=\frac{\frac{2}{7}}{1-\frac{1}{7}}$
$\Rightarrow \frac{36x}{49}-1=\frac{\frac{2}{7}}{\frac{6}{7}}$
$\Rightarrow \frac{36x}{49}=1+\frac{1}{3}=\frac{4}{3}$

$\therefore x=\frac{49}{27}$