Question

2/1x3 2/3x5 2/5x7 ... 2/99x101

sos 

Answer 1

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

Answer 2

=1−13+13−15+...+199−1101=1−13+13−15+...+199−1101