Question

$1+\frac{2}{1.2.3}+\frac{2}{3.4.5}+\frac{2}{5.6.7}+...$ is

• A. $2{\log }_{e}2$
• B. $2{\log }_{e}4$
• C. $2{\log }_{e}3$
• D. $noneofthese$

Answer 1

The correct option is A

$2{\log }_{e}2$

Explanation for the correct option:

Step 1. Rewriting each term of given series as difference of two terms

Let $S=1+\frac{2}{1.2.3}+\frac{2}{3.4.5}+\frac{2}{5.6.7}+...$ as:

We can write $\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}$

Similarly,

$\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}$

$\frac{2}{5.6.7}=\frac{1}{5.6}-\frac{1}{6.7}$ and so on

Step 2. Find the sum by putting these values in given series:

$\Rightarrow S=1+\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+.....$
$\Rightarrow S=1+\left(\frac{1}{1.2}+\frac{1}{3.4}+....\right)-\left(\frac{1}{2.3}+\frac{1}{4.5}+....\right)$
$\Rightarrow S=1+\left(\frac{1}{1.2}+\frac{1}{3.2^2}+....\right)-\left(\frac{1}{2.3}+\frac{1}{4.5}+....\right)$
$\Rightarrow S=1+{\log }_{e}2-\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+.....\right]$ $\left[\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\left(1+x\right)\mathbf{=}\mathit{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}{\mathit{x}}^{\mathbf{3}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\right]$

$\Rightarrow S=1+{\log }_{e}2-1+{\log }_{e}2$

$\Rightarrow S=\mathbf{2}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathbf{2}$

Hence, Option ‘A’ is Correct.