CAT 2003 Slot 2 — QA Question 19

Let S = $1+\frac{4}{7}+\frac{9}{7^2}+\frac{16}{7^3}+\frac{25}{7^4}+.....$ (i)

Dividing (i) by 7, we get,

$\frac{S}{7}=\frac{1}{7}+\frac{4}{7^2}+\frac{9}{7^3}+\frac{16}{7^4}+\frac{25}{7^5}+....$(ii)
Subtracting (ii) from (i)

$\frac{6S}{7}=1+\frac{3}{7}+\frac{5}{7^2}+\frac{7}{7^3}+\frac{9}{7^4}+....$(iii)
Dividing (iii) by 7

$\frac{6S}{49}=\frac{1}{7}+\frac{3}{7^2}+\frac{5}{7^3}+\frac{5}{7^4}+...$(iv)
Subtracting (iv) from (iii)

$\frac{36S}{49}=1+2\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...\right)=1+\frac{2\times {\displaystyle \frac{1}{7}}}{1-{\displaystyle \frac{1}{7}}}=1+\frac{{\displaystyle \frac{2}{7}}}{{\displaystyle \frac{6}{7}}}=\frac{4}{3}$

$$\begin{gathered} \therefore \frac{36S}{49}=\frac{4}{3} \\ S=\frac{4}{3}\times \frac{49}{36}=\frac{49}{27} \end{gathered}$$

Hence, option (c).