Question

\(B=\frac{1}{19}+\frac{9}{19\times29}+\frac{9}{29\times39}+.......+\frac{9}{1999\times2009}\)

Answer 1

\(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)

\(=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)=\frac{9}{10}\cdot\frac{2000}{18081}=\frac{200}{2009}\)

Answer 2

Ta có: \(B=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)

\(B=9\left(\frac{1}{9.19}+\frac{1}{19.29}+...+\frac{1}{1999.2009}\right)\)

\(B=\frac{9}{10}\left(\frac{10}{9.19}+\frac{10}{19.29}+...+\frac{10}{1999.2009}\right)\)

\(B=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(B=\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)

\(B=\frac{9}{10}.\frac{2000}{18081}\)

\(B=\frac{200}{2009}\)

Vậy \(B=\frac{200}{2009}\)