Question

tinh

2003/1.2+2003/2.3+...+2003/2002.2003

Answer 1

\(\frac{2003}{1\cdot2}+\frac{2003}{2\cdot3}+...+\frac{2003}{2002\cdot2003}\)

\(=2003\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2002\cdot2003}\right)\)

\(=2003\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{2002}-\frac{1}{2003}\right)\)

\(=2003\cdot\left(1-\frac{1}{2003}\right)\)

\(=2003\cdot\frac{2002}{2003}\)

\(=\frac{2003\cdot2002}{2003}\)

\(=2002\)