\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+....+\frac{1}{2003\cdot2005}\)
\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2003\cdot2005}\)
\(2A=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{2005-2003}{2003\cdot2005}\)
\(2A=\left(\frac{3}{1\cdot3}-\frac{1}{1\cdot3}\right)+\left(\frac{5}{3\cdot5}-\frac{3}{3\cdot5}\right)+\left(\frac{7}{5\cdot7}-\frac{5}{5\cdot7}\right)+....+\left(\frac{2005}{2003\cdot2005}-\frac{2003}{2003\cdot005}\right)\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2003}-\frac{1}{2005}\)
\(2A=1-\frac{1}{2005}\)
\(2A=\frac{2004}{2005}\)
\(A=\frac{2004}{2005}\div2\)
\(A=\frac{2004}{4010}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)
( GẠCH BỎ CÁC PHÂN SỐ GIỐNG NHAU )
\(=1-\frac{1}{2005}\)
\(=\frac{2004}{2005}\)
CHÚC BN HỌC TỐT!!!
Mấy bạn làm sai rồi nhé =))) Mình từng làm bài này rồi nên biết
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(\Leftrightarrow A=\frac{1}{2}\left(1-\frac{1}{2005}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{2004}{2005}=\frac{1002}{2005}\)
