Question

$\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...is$

• A. $2\log 2-1$
• B. $3\log 2$
• C. $4\log 2–1$
• D. $Noneofthese$

Answer 1

The correct option is A

$2\log 2-1$

Explanation for the correct option:

Step 1. Simplify given series

Let $S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}–\frac{1}{4.5}+..\dots$

$=\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}–\frac{1}{3}\right)+\left(\frac{1}{3}–\frac{1}{4}\right)–\left(\frac{1}{4}–\frac{1}{5}\right)+\dots$

$=\left(1-\frac{1}{2}+\frac{1}{3}–\frac{1}{4}+\frac{1}{5}-\dots \right)+\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots \right)$ ……….(1)

Step 2. Use log expansion formula

As we know, $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}–\frac{x^4}{4}+..$

Put $x=1$ in above equation

$\Rightarrow \log 2=1–\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}....$

$\Rightarrow \log 2-1=–\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}....$

Put the value of $\log 2$ and $\log 2-1$ in equation (1)

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

Hence, Option ‘A’ is Correct.