Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& \frac{1\times 2^2+2\times 3^2+\dots +100\times (101{)}^2}{1^2\times 2+2^2\times 3+\dots +100^2\times 101}=\frac{{\sum }_{r=1}^{100}r(r+1{)}^2}{{\sum }_{r=1}^{100}{r}^2(r+1)}\\ & =\frac{{\sum }_{r=1}^{100}\left(r^3+2r^2+r\right)}{{\sum }_{r=1}^{100}\left(r^3+r^2\right)}=\frac{\left(\frac{n(n+1{)}^2}{2}\right)+\frac{2\cdot n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{{\left(\frac{n(n+1)}{2}\right)}^2+\frac{n(n+1)(2n+1)}{6}}\\ & =\frac{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3}\cdot (2n+1)+1\right]}{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{(2n+1)}{3}\right]};\text{ Put n=100 }\\ & =\frac{\frac{100(101)}{2}+\frac{2}{3}(201)+1}{\frac{100\times 101}{2}+\frac{201}{3}}=\frac{5185}{5117}=\frac{305}{301}\\ & \end{array}$