Question

Tính A:

A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

HELP ME!

Answer 1

A = \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)

Vậy: A = \(\dfrac{49}{100}\)

Answer 2

A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{100}\)

A=\(\dfrac{49}{100}\)

Answer 3

\(\dfrac{49}{100}\)