Question

The sum of the series 11.2−12.3+13.411.2−12.3+13.4 is equal to

• A. $lo{g}_{e}2-1$
• B. $lo{g}_{e}2$
• C. $lo{g}_e\left(\frac{4}{e}\right)$
• D. $2lo{g}_{e}2$

Answer 1

The correct option is A $lo{g}_{e}2-1$

Given,

$\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}+\frac{1}{3+4}-\frac{1}{4\cdot 5}+\cdots$

We can write it as,

$(1-\frac{1}{2})-(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})-(\frac{1}{4}-\frac{1}{5})+-$

$\Rightarrow (1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+1+(-\frac{1}{2}+\frac{1}{3}\frac{1}{4}+\frac{1}{5})$

because we know,

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\cdot \cdot \cdot$

Putting x = 1

We get,

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

Putting the value of log 2 in the above equation,

$\Rightarrow \log 2+\log 2-1$

$\Rightarrow 2\log 2-1=\log (2{)}^2-\log e$

$=\log \left(\frac{4}{e}\right)$

Option C