Question

The sum of the infinite series $1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}+.....$ is equal to

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{\frac{1}{3}}$

Answer 1

The correct option is B $\sqrt{3}$

Given series,
$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}+.....\infty$

$=1+\frac{1}{3}+\frac{\frac{1}{2}\cdot \frac{3}{2}}{1\cdot 2}{\left(\frac{2}{3}\right)}^2+\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot \frac{5}{2}}{1\cdot 2\cdot 3}{\left(\frac{2}{3}\right)}^3.....\infty$

$=1+\frac{1}{2}\cdot \frac{2}{3}+\frac{\frac{1}{2}(\frac{1}{2}+1)}{2!}{\left(\frac{2}{3}\right)}^2+\frac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)}{3!}{\left(\frac{2}{3}\right)}^3......\infty$

$={(1-\frac{2}{3})}^{-1/2}=3^{1/2}=\sqrt{3}$