Question

Tính nhanh:

\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2001.2003}+\frac{2}{2003.2005}\)

Answer 1

Ta có: 

A =  \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\) + ... + \(\frac{2}{2001.2003}\) + \(\frac{2}{2003.2005}\)

    =  \(\frac{1}{1}\) -  \(\frac{1}{3}\)+ \(\frac{1}{3}\)-  \(\frac{1}{5}\)+ \(\frac{1}{5}\)-  \(\frac{1}{7}\)+  \(\frac{1}{7}\)-  \(\frac{1}{9}\) + ... + \(\frac{1}{2001}\)- \(\frac{1}{2003}\)+ \(\frac{1}{2003}\)-  \(\frac{1}{2005}\)

    = 1 - \(\frac{1}{2005}\)

     = \(\frac{2004}{2005}\)

Chúc bạn học tốt nha ^^!!