Question

B=1/1.3+1/3.5+...+1/2017.2019

Answer 1

\(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2017.2019}\)

\(2B=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2B=\frac{1}{1}-\frac{1}{2019}\)

\(2B=\frac{2018}{2019}\)

\(\Rightarrow B=\frac{2018}{2019}:2\Rightarrow B=\frac{1009}{2019}\)

Answer 2

\(B=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{2}{2017.2019}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{2017}-\frac{1}{2019}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2019}\right)=\frac{1009}{2019}\)