Question

Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\)

 \(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{2}{2010}+\frac{1}{2011}\)

Tính \(\frac{A}{B}\)

Answer 1

Có B = \(\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{1}{2011}\)

B = \(\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+....+\left(\frac{1}{2011}+1\right)+1\)

B = \(\frac{2012}{2}+\frac{2012}{3}+....+\frac{2012}{2011}+\frac{2012}{2012}\)

B = \(2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)\)

=> \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}=\frac{1}{2012}\)