Ta thấy:
\(1+2=\dfrac{2\cdot\left(2+1\right)}{2}\\ 1+2+3=\dfrac{3\cdot\left(3+1\right)}{2}\\ 1+2+3+4=\dfrac{4\cdot\left(4+1\right)}{2}\\ ...\\ 1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\\ =1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{20}\cdot\dfrac{20\cdot21}{2}\\ =1+\dfrac{1\cdot2\cdot3}{2\cdot2}+\dfrac{1\cdot3\cdot4}{3\cdot2}+...+\dfrac{1\cdot20\cdot21}{20\cdot2}\\ =\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\\ =\dfrac{2+3+4+...+21}{2}\\ =\dfrac{1+2+3+..+21-1}{2}\\ =\dfrac{\left(\dfrac{21\cdot22}{2}\right)-1}{2}\\ =\dfrac{231-1}{2}\\ =\dfrac{230}{2}\\ =115\)