Question

Tính:

A=\(\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2014.2016}\right)\)

Lưu ý: Dấu . là dấu nhân

Answer 1

\(A=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{24}\right).\left(1+\dfrac{1}{3.5}\right).....\left(1+\dfrac{1}{2014.2016}\right)\)

\(A=\dfrac{4}{1.3}.\dfrac{9}{2.4}.\dfrac{16}{3.5}.....\dfrac{4060225}{2014.2016}\)

\(A=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{2015^2}{2014.2016}\)

\(A=\dfrac{2.3.4.5...2015}{1.2.3...2014}.\dfrac{2.3.4...2015}{3.4.5...2016}\)

\(A=2015.\dfrac{2}{2016}=2015.\dfrac{1}{1008}=\dfrac{2015}{1008}\)

Vậy \(A=\dfrac{2015}{1008}\)