Question

A = (1+ 1/1x3) x ( 1 + 1/2x4) + ..... + (1 + 1/2018x2020)

Tính A

Answer 1

Ta có: \(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2018\cdot2020}\right)\)

\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{2019^2-1}\right)\)

\(=\frac{2^2-1+1}{2^2-1}\cdot\frac{3^2-1+1}{3^2-1}\cdot\ldots\cdot\frac{2019^2-1+1}{2019^2-1}\)

\(=\frac{2^2}{2^2-1}\cdot\frac{3^2}{3^2-1}\cdot...\cdot\frac{2019^2}{2019^2-1}\)

\(=\frac{2\cdot3\cdot\ldots\cdot2019}{1\cdot2\cdot\ldots\cdot2018}\cdot\frac{2\cdot3\cdot\ldots\cdot2019}{3\cdot4\cdot\ldots\cdot2020}=\frac{2019}{1}\cdot\frac{2}{2020}=\frac{2019}{1010}\)