Question

Tìm tổng:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}+...+\dfrac{1}{255.257}\)

Ta được kết quả là:

A.\(\dfrac{127}{255}\) B.\(\dfrac{128}{255}\) C.\(\dfrac{128}{257}\) D.\(\dfrac{129}{257}\)

Chọn đáp án rồi giải ra giúp mik với nha các bạn!Mik cảm ơn trước.

Answer 1

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1}{255\cdot257}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{2}{255\cdot257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}+...+\dfrac{1}{255}-\dfrac{1}{257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{257}\right)\)\(=\dfrac{1}{2}\cdot\dfrac{256}{257}=\dfrac{128}{257}\)