Question

\(\dfrac{3}{1×3}\)+\(\dfrac{3}{3×5}\)+.........+\(\dfrac{3}{97×99}\)

Answer 1

\(\dfrac{3}{1.3}+\dfrac{3}{1.5}+...+\dfrac{3}{97.99}\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{97.99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\dfrac{98}{99}\)

\(=\dfrac{1}{1}.\dfrac{49}{33}\)

\(=\dfrac{49}{33}\)

Answer 2

= 1 - 1/3 + 1/ 3 - 1/5 + 1/ 5 - ... + 1/97 -1/99

=1-1/99

=98/99

Answer 3

\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+...+\dfrac{3}{97.99}\)

\(=3.\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\dfrac{98}{99}\)

\(=\dfrac{49}{33}\)

Answer 4

\(\dfrac{3}{1\cdot3}+\dfrac{3}{3\cdot5}+...+\dfrac{3}{97\cdot99}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{3}{2}\cdot\dfrac{98}{99}=\dfrac{49}{33}\)