Note: It is known that ${\displaystyle \underset{i=1}{\overset{n}{}}}i=\frac{n(n+1)}{2}.$ Let $n0,1f(x)=x{R}_n={\displaystyle \underset{i=1}{\overset{n}{}}}\frac{i}{n}=\frac{n(n+1)}{2n}{R}_n={\displaystyle \underset{i=1}{\overset{n}{}}}\frac{i}{n}*\frac{1}{n}=\frac{n(n+1)}{2n^2}{R}_n={\displaystyle \underset{i=1}{\overset{n}{}}}\frac{i-1}{n}*\frac{1}{n}=\frac{(n-1)n}{2n^2}$