Question

Tính tổng đẳng thức sau

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

Answer 1

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

\(2A=2.\left(\frac{1}{1.3}+\frac{1}{2.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\right)\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2003.2005}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\)

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

\(2A=\frac{2004}{2005}\)

\(A=\frac{2004}{2005}:2\)

\(A=\frac{1002}{2005}\)

Ủng hộ tk Đúng nha mọi người !!! ^^ 

Answer 2

Đặt B = \(\frac{1}{1.3}\)+ \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\Rightarrow2B=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\right)\)\(\Rightarrow2B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2003.2005}\Rightarrow2B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2003}-\frac{1}{2005}\)

\(\Rightarrow2B=\frac{1}{3}-\frac{1}{2005}=\frac{2012}{6015}\Rightarrow B=\frac{2012}{6015}:2=\frac{1001}{6015}\)

Answer 3

Ta có: \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{2003.2005}\)

             \(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{2003.2005}\right)\)

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

               \(=\frac{1}{2}\left(1-\frac{1}{2005}\right)\)

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