The value of $\frac{1 imes 2^2+2 imes 3^2+\ldots+100 imes(101)^2}{1^2 imes 2+2^2 imes 3+\ldots .+100^2 imes 101}$ is

Question

The value of $\frac{1 	imes 2^2+2 	imes 3^2+\ldots+100 	imes(101)^2}{1^2 	imes 2+2^2 	imes 3+\ldots .+100^2 	imes 101}$ is

Aicharya · Accepted Answer

$\begin{aligned} & \frac{1 	imes 2^2+2 	imes 3^2+\ldots+100 	imes(101)^2}{1^2 	imes 2+2^2 	imes 3+\ldots+100^2 	imes 101}=\frac{\sum_{r=1}^{100} r(r+1)^2}{\sum_{r=1}^{100} r^2(r+1)} \ & =\frac{\sum_{r=1}^{100}\left(r^3+2 r^2+right)}{\sum_{r=1}^{100}\left(r^3+r^2ight)}=\frac{\left(\frac{n(n+1)^2}{2}ight)+\frac{2 \cdot n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}ight)^2+\frac{n(n+1)(2 n+1)}{6}} \ & =\frac{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3} \cdot(2 n+1)+1ight]}{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{(2 n+1)}{3}ight]} ; 	ext { Put n=100 } \ & =\frac{\frac{100(101)}{2}+\frac{2}{3}(201)+1}{\frac{100 	imes 101}{2}+\frac{201}{3}}=\frac{5185}{5117}=\frac{305}{301} \ & \end{aligned}$

The value of $\frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots …

Solution