$\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+ \ldots \ldots \ldots+\dfrac{1}{90}$
Given series can be written like this:
$\dfrac{1}{2*3}+\dfrac{1}{3*4}+\dfrac{1}{4*5}+\ldots\ldots+\dfrac{1}{9*10}$
The above series can be written as:
$\implies\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dots\ldots+\dfrac{1}{9}-\dfrac{1}{10}$
$\implies \dfrac{1}{2}-\dfrac{1}{10}$
$\implies \dfrac{5-1}{10}=\dfrac{4}{10}=\dfrac{2}{5}$
Option $(B)$ is correct.