Q49 / 90 · JEE Main 2024 (06 Apr Shift 2) · Mathematics

$\lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)}$ is equal to :

  • A
    $\frac{2}{3}$
  • Correct answer
    B
    $\frac{1}{3}$
  • C
    $\frac{3}{4}$
  • D
    $\frac{1}{2}$

Solution

$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{\sum_{r=1}^n r^3-\sum_{r=1}^n r^2} \\ & \lim _{\mathrm{n} \rightarrow \infty} \frac{\sum_{\mathrm{r}=1}^{\mathrm{n}-1}\left(-\mathrm{r}^3+\mathrm{r}^2(\mathrm{n}+1)-\mathrm{nr}\right)}{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^2-\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}} \\ & \lim _{n \rightarrow \infty} \frac{\left(\frac{((n-1) n)}{2}\right)^2+\frac{(n+1)(n-1) n(2 n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)} \\ & \lim _{n \rightarrow \infty} \frac{\frac{n(n-1)}{2}\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2 n-1)}{3}-n\right)}{\frac{n(n+1)}{2} \frac{3 n^2+3 n-4 n-2}{6}} \\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(-3 n^2+3 n+2\left(2 n^2+n-1\right)-6\right)}{(n+1)\left(3 n^2-n-2\right)} \\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(n^2+5 n-8\right)}{(n+1)\left(3 n^2-n-2\right)}=\frac{1}{3} \\ & \end{aligned}$

Hence, the correct answer is option: 'B'.

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