Question

\(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) +......+ \(\dfrac{1}{(2x-1)(2x+1)}\) = \(\dfrac{49}{99}\)

HeLp me

Answer 1

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)

=> 2x=98

=> x=49