Question

2/2.3 + 2/3.4 +...+ 2/99.100

help :C

Answer 1

\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\)

\(=\)\(\dfrac{1\cdot2}{2\cdot3}+\dfrac{1\cdot2}{3\cdot4}+...+\dfrac{1\cdot2}{99\cdot100}\)

\(=\)\(2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)

\(=\)\(2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\)\(2\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=\)\(2\cdot\dfrac{49}{100}\)

\(=\)\(\dfrac{49}{50}\)

Answer 2

= 1/1 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

= 1/1 - 1/100

= 99/100