JEE Main 2024 — Mathematics Algebra

$$\begin{aligned} & \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101}=\frac{\sum_{r=1}^{100} r(r+1)^2}{\sum_{r=1}^{100} r^2(r+1)} \\ & =\frac{\sum_{r=1}^{100}\left(r^3+2 r^2+r\right)}{\sum_{r=1}^{100}\left(r^3+r^2\right)}=\frac{\left(\frac{n(n+1)^2}{2}\right)+\frac{2 \cdot n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}}{\left(\frac{n(n+1)}{2}\right)^2+\frac{n(n+1)(2 n+1)}{6}} \\ & =\frac{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3} \cdot(2 n+1)+1\right]}{\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{(2 n+1)}{3}\right]} ; \text { Put n=100 } \\ & =\frac{\frac{100(101)}{2}+\frac{2}{3}(201)+1}{\frac{100 \times 101}{2}+\frac{201}{3}}=\frac{5185}{5117}=\frac{305}{301} \\ & \end{aligned}$$​12×2+22×3+…+1002×1011×22+2×32+…+100×(101)2​=∑r=1100​r2(r+1)∑r=1100​r(r+1)2​=∑r=1100​(r3+r2)∑r=1100​(r3+2r2+r)​=(2n(n+1)​)2+6n(n+1)(2n+1)​(2n(n+1)2​)+62⋅n(n+1)(2n+1)​+2n(n+1)​​=2n(n+1)​[2n(n+1)​+3(2n+1)​]2n(n+1)​[2n(n+1)​+32​⋅(2n+1)+1]​; Put n=100 =2100×101​+3201​2100(101)​+32​(201)+1​=51175185​=301305​​