Question

tính:1/4x9 + 1/9x4 + 1/14x19 + ... + 1/1999x2004

Answer 1

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{1999\cdot2004}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}\cdot\dfrac{125}{501}=\dfrac{25}{501}\)

Answer 2

Sửa \(\dfrac{1}{9.4}\rightarrow\dfrac{1}{9.14}\) nha

\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+...+\dfrac{5}{1999.2004}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}.\dfrac{125}{501}=\dfrac{25}{501}\)

Answer 3

\(=2\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+.....+\dfrac{1}{1999.2004}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=2\cdot\dfrac{125}{501}=\dfrac{250}{501}\)