Question

K=\(\dfrac{4}{1x3}\)+\(\dfrac{4}{3x5}\)+\(\dfrac{4}{5x7}\)+...+\(\dfrac{4}{297x299}\)+\(\dfrac{4}{299x301}\)

Answer 1

\(K=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+...+\dfrac{4}{299\times301}\)

\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{299\times301}\right)\)

\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)

\(=2\times\left(1-\dfrac{1}{301}\right)=2\times\dfrac{300}{301}=\dfrac{600}{301}\)

Answer 2

\(K=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+...+\dfrac{4}{299\cdot301}\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)

\(=2\cdot\dfrac{300}{301}=\dfrac{600}{301}\)