Question

The value of $1+\frac{1}{3}+\frac{1·3}{3·6}+\frac{1·3·5}{3·6·9}+\frac{1·3·5·7}{3·6·9·12}+............$ is equal to

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

Answer 1

The correct option is B

$\sqrt{3}$

Explanation for the correct option.

Find the value.

Given: $1+\frac{1}{3}+\frac{1·3}{3·6}+\frac{1·3·5}{3·6·9}+\frac{1·3·5·7}{3·6·9·12}+............$

Rewrite the above expression

$=1+\left(\frac{1}{2}\right)·\left(\frac{2}{3}\right)+\left[\left(\frac{1}{2}\right).\left(\frac{3}{2}\right)·{\left(\frac{2}{3}\right)}^2·\frac{1}{2!}\right]+\left[\left(\frac{1}{2}\right)·\left(\frac{3}{2}\right)·\left(\frac{5}{2}\right)·{\left(\frac{2}{3}\right)}^3·\frac{1}{3!}\right]+.........-\left(1\right)$

Divide and multiply 2nd term by$2$, and divide and multiply 3rd term by $4$and so on increasing in multiple of $2$

${\left(1-\mathrm{x}\right)}^{-\mathrm{n}}=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}+1)}{1·2}{\mathrm{x}}^2+\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{1·2·3}{\mathrm{x}}^3+.......$

Equation (1) is the same as the above binomial series where $\mathrm{n}=\frac{1}{2},\mathrm{x}=\frac{2}{3}$

Therefore equation (1) is equal to ${\left(1-\frac{2}{3}\right)}^{-\frac{1}{2}}$

$\begin{array}{rcl}{\left(1-\frac{2}{3}\right)}^{-\frac{1}{2}}& =& {\left(\frac{3-2}{3}\right)}^{-\frac{1}{2}}\\ & =& {\left(\frac{1}{3}\right)}^{-\frac{1}{2}}\\ & =& {\left(3\right)}^{\frac{1}{2}}\\ & =& \sqrt{3}\end{array}$

Hence $o\mathrm{ption}\left(\mathrm{B}\right)$ is the correct answer.