Question

Tính tổng sau:

A = \(\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)

help me!

Answer 1

\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)

Answer 2

\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)

\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)