JEE Main 2024 — Mathematics Calculus

$$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{\sum_{r=1}^n r^3-\sum_{r=1}^n r^2} \\ & \lim _{\mathrm{n} \rightarrow \infty} \frac{\sum_{\mathrm{r}=1}^{\mathrm{n}-1}\left(-\mathrm{r}^3+\mathrm{r}^2(\mathrm{n}+1)-\mathrm{nr}\right)}{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^2-\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}} \\ & \lim _{n \rightarrow \infty} \frac{\left(\frac{((n-1) n)}{2}\right)^2+\frac{(n+1)(n-1) n(2 n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)} \\ & \lim _{n \rightarrow \infty} \frac{\frac{n(n-1)}{2}\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2 n-1)}{3}-n\right)}{\frac{n(n+1)}{2} \frac{3 n^2+3 n-4 n-2}{6}} \\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(-3 n^2+3 n+2\left(2 n^2+n-1\right)-6\right)}{(n+1)\left(3 n^2-n-2\right)} \\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(n^2+5 n-8\right)}{(n+1)\left(3 n^2-n-2\right)}=\frac{1}{3} \\ & \end{aligned}$$​n→∞lim​∑r=1n​r3−∑r=1n​r2∑r=1n−1​(r2−r)(n−r)​n→∞lim​(2n(n+1)​)2−6n(n+1)(2n+1)​∑r=1n−1​(−r3+r2(n+1)−nr)​n→∞lim​2n(n+1)​(2n(n+1)​−32n+1​)(2((n−1)n)​)2+6(n+1)(n−1)n(2n−1)​−2n2(n−1)​​n→∞lim​2n(n+1)​63n2+3n−4n−2​2n(n−1)​(2−n(n−1)​+3(n+1)(2n−1)​−n)​n→∞lim​(n+1)(3n2−n−2)(n−1)(−3n2+3n+2(2n2+n−1)−6)​n→∞lim​(n+1)(3n2−n−2)(n−1)(n2+5n−8)​=31​​