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$\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+ \ldots \ldots \ldots+\dfrac{1}{90}$

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.
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