Question

A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}\)

B = \(\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)

Tính \(\dfrac{B}{A}\)

Answer 1

Đặt: \(L_2=\dfrac{2007}{1}+\dfrac{2006}{2}+\dfrac{2005}{3}+...+\dfrac{2}{2006}+\dfrac{1}{2007}\)

\(L_2=1+\left(\dfrac{2006}{2}+1\right)+\left(\dfrac{2005}{3}+1\right)+...+\left(\dfrac{2}{2006}+1\right)+\left(\dfrac{1}{2007}+1\right)\)

\(L_2=\dfrac{2008}{2008}+\dfrac{2008}{2}+\dfrac{2008}{3}+...+\dfrac{2008}{2006}+\dfrac{2008}{2007}\)

\(L_2=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+..+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

\(\dfrac{L_1}{L_2}=\dfrac{1}{2008}\)