$\lim_{n \rightarrow \infty} \frac{\left(1^{2}-1\right)(n-1)+\left(2^{2}-2\right)(n-2)+\ldots\ldots\ldots+\left((n-1)^{2}-(n-1)\right) \cdot 1}{\left(1^{3}+2^{3}+\ldots+n^{2}\right)-\left(1^{2}+2^{2}+\ldots+n^{2}\right)}$ is equal to

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Limits

lim_{n \to \infty} \frac{( 1 ^{2} - 1 ) ( n - 1 ) + ( 2 ^{2} - 2 ) ( n - 2 ) + \dots\dots\dots + ( ( n - 1 ) ^{2} - ( n - 1 ) ) \cdot 1}{( 1 ^{3} + 2 ^{3} + \dots + n ^{2} ) - ( 1 ^{2} + 2 ^{2} + \dots + n ^{2} )}

is equal to

Answer: (C)

Both series are quartic in

n

. Using

\sum r \sim \frac{n ^{2}}{2}

\sum r^{2} \sim \frac{n ^{3}}{3}

\sum r^{3} \sim \frac{n ^{4}}{4}

: the numerator

r = 1 \sum n - 1 (r^{2} - r) (n - r) = \sum [(n + 1) r^{2} - r^{3} - n r] \sim n \cdot \frac{n ^{3}}{3} - \frac{n ^{4}}{4} = \frac{n ^{4}}{12}

, while the denominator

r = 1 \sum n r^{3} - r = 1 \sum n r^{2} \sim \frac{n ^{4}}{4}

. The limit is

\frac{n ^{4} /12}{n ^{4} /4} = \frac{1}{3}

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Single correct option