\(a,=\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right]\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
\(b,=\dfrac{1}{4}\left[\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right]\\ =\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\\ =\dfrac{1}{4}\left(1-\dfrac{1}{4n+1}\right)=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
