Question

A= 4/2x3 + 4/3x4 + 4/4x5 + ... + 4/99x100

 

Answer 1

\(\dfrac{4}{2\times3}+\dfrac{4}{3\times4}+\dfrac{4}{4\times5}+...+\dfrac{4}{99\times100}\)
\(=4\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=4\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(=4\left(\dfrac{50}{100}-\dfrac{1}{100}\right)\)
\(=4\times\dfrac{49}{100}\)
\(=\dfrac{49}{25}\)