Question

$\frac{1}{1.3}+\frac{1}{2.5}+\frac{1}{3.7}+\frac{1}{4.9}+\dots \dots =$

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

Answer 1

The correct option is B $2-{\log }_{e}2$

The general term for the above expression is
$T_n=\frac{1}{\left(n\right)(2n+1)}$
$=\frac{2n+1-2n}{\left(n\right)(2n+1)}$
$=\frac{1}{n}-\frac{2}{(2n+1)}$
Hence
$S=1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}..$
$=(1+\frac{1}{2}+\frac{1}{3}+....)-(\frac{2}{3}+\frac{2}{5}+\frac{2}{7}....)$
$=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...$
$=2+(-1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...)$
$=2-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+...)$
$=2-\log 2$