Question

Tính A biết :

A=\(\dfrac{1}{1.3}+\)\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+....+\(\dfrac{1}{2003.2005}\)

MỌI NGƯỜI GIÚP MÌNH VỚI Ạ!

 

Answer 1

\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2003\cdot2005}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2003\cdot2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2005}\right)=\dfrac{1}{2}\cdot\dfrac{2004}{2005}=\dfrac{1002}{2005}\)

Answer 2

\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}\)

\(A=\dfrac{2004}{2005}\).

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