QJEEM24Ma2358: JEE Main 2024 Mathematics - Class 12 JEEM24Ma2358 Solution: lim n → ∞ ( 1 2 − 1 ) ( n − 1 ) + ( 2 2 − 2 ) ( n − 2 ) + ⋯ + ( …

$\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}$

$\begin{aligned}& \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \\& =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \\& =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n\end{aligned}$

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 $\frac{n^4}{4}$

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{12}{n}^4}{\frac{n^4}{4}}=\frac{1}{3}$$