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Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 6 April, Shift 2 — Question 6

lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+….+((n−1)2−(n−1))⋅1(13+23+….+n3)−(12+22+….+n2)\quad \lim _{n \rightarrow \infty} \frac{\left(1^{2}-1\right)(n-1)+\left(2^{2}-2\right)(n-2)+\ldots .+\left((n-1)^{2}-(n-1)\right) \cdot 1}{\left(1^{3}+2^{3}+\ldots .+n^{3}\right)-\left(1^{2}+2^{2}+\ldots .+n^{2}\right)} is equal to:

  1. Option A:

    23\frac{2}{3}

  2. Option B:

    13\frac{1}{3}

    Correct
  3. Option C:

    34\frac{3}{4}

  4. Option D:

    12\frac{1}{2}

Answer: B

Step-by-step solution

lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1(13+23+⋯+n3)−(12+22+⋯+n2)\lim_{n \to \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+n^{3}\right)-\left(1^{2}+2^{2}+\cdots+n^{2}\right)}

Simplify numerator

The numerator is

∑k=1n−1(k2−k)(n−k)=∑k=1n−1k(k−1)(n−k)\sum_{k=1}^{n-1} (k^2-k)(n-k) = \sum_{k=1}^{n-1} k(k-1)(n-k)

Expand:

k(k−1)(n−k)=(n+1)k2−nk−k3k(k-1)(n-k) = (n+1)k^2 - nk - k^3

Hence numerator

=(n+1)∑k2−n∑k−∑k3= (n+1)\sum k^2 - n\sum k - \sum k^3

Using standard sums (up to leading terms):

∑k=1n−1k∼n22,∑k2∼n33,∑k3∼n44\sum_{k=1}^{n-1} k \sim \frac{n^2}{2}, \quad \sum k^2 \sim \frac{n^3}{3}, \quad \sum k^3 \sim \frac{n^4}{4}

So leading term:

(n+1)n33−nn22−n44(n+1)\frac{n^3}{3} - n\frac{n^2}{2} - \frac{n^4}{4} =n43−n32−n44= \frac{n^4}{3} - \frac{n^3}{2} - \frac{n^4}{4} =n412+O(n3)= \frac{n^4}{12} + O(n^3)

Denominator

∑k=1nk3−∑k=1nk2\sum_{k=1}^n k^3 - \sum_{k=1}^n k^2 ∑k3=n2(n+1)24∼n44\sum k^3 = \frac{n^2(n+1)^2}{4} \sim \frac{n^4}{4} ∑k2=n(n+1)(2n+1)6∼n33\sum k^2 = \frac{n(n+1)(2n+1)}{6} \sim \frac{n^3}{3}

Thus leading term:

n44−n33=n44+O(n3)\frac{n^4}{4} - \frac{n^3}{3} = \frac{n^4}{4} + O(n^3)

Take limit

lim⁡n→∞n412n44=1/121/4=412=13\lim_{n\to\infty} \frac{\frac{n^4}{12}}{\frac{n^4}{4}} = \frac{1/12}{1/4} = \frac{4}{12} = \frac13 13\boxed{\frac13}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 2
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Indeterminate forms & its solving methods

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