Question

Tính p=1/2+1/3+1/4+..........+1/2005

Phần 2004/1+2003/1+.......+1/2004

Answer 1

\(P=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.....+\dfrac{1}{2005}}{\dfrac{2004}{1}+\dfrac{2003}{2}+....+\dfrac{1}{2004}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2004}+\dfrac{1}{2005}}{\left(\dfrac{2003}{2}+1\right)+\left(\dfrac{2002}{3}+1\right)+.....+\left(\dfrac{1}{2004}+1\right)}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{2005}{2}+\dfrac{2005}{3}+.....+\dfrac{2005}{2005}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2004}+\dfrac{1}{2005}}{2005\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2005}\right)}\)

\(=\dfrac{1}{2005}\)