JEE Main 2024 — 4 April, Shift 2 — Question 7

$\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+100 \times(101)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+100^{2} \times 101}=\frac{\sum_{r=1}^{100} r(r+1)^{2}}{\sum_{r=1}^{100} r^{2}(r+1)}$

$=\frac{\sum_{\mathrm{r}=1}^{100}\left(r^{3}+2 r^{2}+r\right)}{\sum_{\mathrm{r}=1}^{100}\left(\mathrm{r}^{3}+r^{2}\right)}=\frac{\left(\frac{\mathrm{n}(\mathrm{n}+1)^{2}}{2}\right)+\frac{2 . \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}+\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}}$

$=\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{2}{3} \cdot(2 \mathrm{n}+1)+1\right]}{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{(2 \mathrm{n}+1)}{3}\right]} ;$ Put $\mathrm{n}=100$

$=\frac{\frac{100(101)}{2}+\frac{2}{3}(201)+1}{\frac{100 \times 101}{2}+\frac{201}{3}}=\frac{5185}{5117}=\frac{305}{301}$