Question

\(A=\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+\dfrac{4}{11}.\dfrac{4}{15}+...+\dfrac{4}{95}.\dfrac{4}{99}\)

Answer 1

128/99

Answer 2

A = \(\dfrac{4}{3}\) . \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) . \(\dfrac{4}{11}\) + \(\dfrac{4}{11}\) . \(\dfrac{4}{15}\) + ... + \(\dfrac{4}{95}\) . \(\dfrac{4}{99}\)

A = \(\dfrac{4.4}{3.7}\) + \(\dfrac{4.4}{7.11}\) + \(\dfrac{4.4}{11.15}\) + ... + \(\dfrac{4.4}{95.99}\)

A = \(\dfrac{16}{3.7}\) + \(\dfrac{16}{7.11}\) + \(\dfrac{16}{11.15}\) + ... + \(\dfrac{16}{95.99}\)

A = 4.( \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) + ... + \(\dfrac{4}{95.99}\))

A = 4.( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\))

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

A = 4.(\(\dfrac{33}{99}\) + \(\dfrac{-1}{99}\))

A = 4. \(\dfrac{32}{99}\)

A = \(\dfrac{4.32}{99}\)

A = \(\dfrac{128}{99}\)

Vậy A = \(\dfrac{128}{99}\)

Answer 3

128/99