Question

\(\frac{1}{1\cdot3}\)+\(\frac{1}{3\cdot5}\)+\(\frac{1}{5\cdot7}\)+......+\(\frac{1}{2003\cdot2005}\)

Answer 1

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

\(=\frac{1}{2}\left[\left(\frac{2005}{2005}-\frac{1}{2005}\right)+0+...+0\right]=\frac{1}{2}\cdot\frac{2004}{2005}=\frac{1002}{2005}\)

Answer 2

Đặt A=\(\frac{1}{1.3}\) + \(\frac{1}{3.5}\) + ... + \(\frac{1}{2003.2005}\)

A= ( 1- \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) +...+ \(\frac{1}{2003}\) - \(\frac{1}{2005}\) ). \(\frac{1}{2}\)

A= ( 1+ \(\frac{1}{3}\) - \(\frac{1}{3}\) + \(\frac{1}{5}\) - \(\frac{1}{5}\) +...+ \(\frac{1}{2003}\) - \(\frac{1}{2003}\) - \(\frac{1}{2005}\) ) .\(\frac{1}{2}\)

A= ( 1- \(\frac{1}{2005}\)).\(\frac{1}{2}\)

A= \(\frac{2004}{2005}\). \(\frac{1}{2}\)= \(\frac{1002}{2005}\)

Answer 3

cảm ơn đã trả lời câu hỏi này cho mk nha

 

Answer 4

1002/2005