JEE MAIN - Mathematics (2024 - 6th April Evening Shift - No. 5)
Explanation
$$\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}$$
$$\begin{aligned} & \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\ & =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \\ & =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n \end{aligned}$$
We will take term with the greatest power of $$n$$
$$=\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4$$
$$\begin{aligned} & \text { Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \\ & =\left(\frac{n(n+1)}{2}\right)^2-\left(\frac{n(n+1)(2 n+1)}{6}\right) \end{aligned}$$
Greatest power of $$n$$ is $$\frac{n^4}{4}$$
$$\lim _\limits{n \rightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}$$
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