Question

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}\dfrac{1}{5\cdot7}+...+\dfrac{1}{2009\cdot2011}\)

Answer 1

`A=1/(1.3)+1/(3.5)+....+1/(2009.2011)`

`=> 2A=2/(1.3)+1/(3.5)+...+2/(2009.2011)`

`=1-1/3+1/3+1/5+.....+1/2009-1/2011`

`=1-1/2011`

`=2010/2011`

`=> A=1005/2011`

Answer 2

Dựa trên công thức: \(\dfrac{a}{n.\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\), ta có:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)

= \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

=\(\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{2010}{2011}\right)\)

= \(\dfrac{2010}{4022}\)

= tự rút gọn nhé

Hok tốt!