Question

$1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+....\infty$ is equal to
[DCE 1999]

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

The correct option is B 6

It is an arithmetico-geometric series
$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}=\frac{1}{1-\frac{1}{2}}+\frac{1}{{(1-\frac{1}{2})}^2}=\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{4}}$
= 2 + 4 = 6.