Question

Tính tổng:

1/1.3+1/3.5+1/5.7+....+1/2017.2019

Answer 1

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(=1-\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2017}-\frac{1}{2019}\div2\)

\(=\left(1-\frac{1}{2019}\right)\div2\)

\(=\frac{2018}{2019}\div2\)

\(=\frac{1009}{2019}\)

Answer 2

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2A=1-\frac{1}{2017}\)

\(2A=\frac{2016}{2017}\)

\(A=\frac{2016}{2017}:2\)

\(A=\frac{1008}{2017}\)

Answer 3

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