Xét tử:
\(\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+....+\frac{19}{1}\)
= \(\left(1+\frac{1}{19}\right)+\left(1+\frac{2}{18}\right)+\left(1+\frac{3}{17}\right)+.....+\left(1+\frac{18}{2}\right)+1\)
= \(\frac{20}{19}+\frac{20}{18}+\frac{20}{17}+.....+\frac{20}{2}+1\)
= \(\frac{20}{20}+\frac{20}{19}+\frac{20}{18}+\frac{20}{17}+...+\frac{20}{2}\)
= \(20\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\right)\)
Thay vào, ta có:
D = \(\frac{20\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}\)
=> D = 20