Question

[Đặng Bình Giang]

Tinh

F=\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)có lời giải nhé😘😘😘

Answer 1

[Kuroba Kaito]

F = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

F = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

F = \(\frac{1}{3}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{9}-\frac{1}{9}\right)-...-\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

F = \(\frac{1}{3}-\frac{1}{99}\)

F = \(\frac{32}{99}\)

Answer 2

[zZz Cool Kid_new zZz]

\(F=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)

\(\Rightarrow F=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\)

\(\Rightarrow F=\frac{1}{3}-\frac{1}{99}\)

\(\Rightarrow F=\frac{32}{99}\)