Question

The sum of the infinite series, $1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+\cdots$, is equal to

• 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 $S=1+\frac{4}{7}+\frac{9}{49}+\frac{16}{343}+......$
$\Rightarrow \frac{S}{7}=\frac{1}{7}+\frac{4}{49}+\frac{9}{343}+.....$
Subtracting them, we get
$\Rightarrow \frac{6S}{7}=1+\frac{3}{7}+\frac{5}{49}+\frac{7}{343}+.....$
Hence, $\frac{6S}{49}=\frac{1}{7}+\frac{3}{49}+\frac{5}{343}+.....$
Subtracting them, we get
$\Rightarrow \frac{36S}{49}=1+\frac{2}{7}+\frac{2}{49}+\frac{2}{343}+....$
$\Rightarrow \frac{36S}{49}=1+\frac{2/7}{6/7}=1+\frac{1}{3}=\frac{4}{3}$
$\Rightarrow S=\frac{49}{27}$