$n=\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+...+\dfrac{2}{99.100}$
$n=\dfrac{2}{3}-\dfrac{2}{4}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{99}-\dfrac{2}{100}$
$n=\dfrac{2}{3}-\dfrac{2}{100}$
$n=\dfrac{200}{300}-\dfrac{6}{300}=\dfrac{194}{300}=\dfrac{97}{150}$