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JEE Mathematics 2024 Question with Solution

If lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+…+((n−1)2−(n−1))(13+23+…+n3)−(12+22+…+n2)\lim_{n\to\infty} \frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\ldots+((n-1)^2-(n-1))}{(1^3+2^3+\ldots+n^3)-(1^2+2^2+\ldots+n^2)} is equal to:

  • A

    23\frac{2}{3}

  • B

    13\frac{1}{3}

  • C

    34\frac{3}{4}

  • D

    12\frac{1}{2}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

L=lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+…+((n−1)2−(n−1))(13+23+…+n3)−(12+22+…+n2)L=\lim_{n\to\infty} \frac{(1^2-1)(n-1)+(2^2-2)(n-2)+\ldots+((n-1)^2-(n-1))}{(1^3+2^3+\ldots+n^3)-(1^2+2^2+\ldots+n^2)}

Find: The value of the limit and hence the correct option.

The solution is unrelated to this question, so the answer is derived from the given answer key.

Write the numerator as

∑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=1n−1(k2−k)(n−k)=n∑k=1n−1(k2−k)−∑k=1n−1(k3−k2)\sum_{k=1}^{n-1}(k^2-k)(n-k)=n\sum_{k=1}^{n-1}(k^2-k)-\sum_{k=1}^{n-1}(k^3-k^2)

Using

∑k=1n−1k=(n−1)n2,∑k=1n−1k2=(n−1)n(2n−1)6,∑k=1n−1k3=((n−1)n2)2\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}, \qquad \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}, \qquad \sum_{k=1}^{n-1} k^3 = \left(\frac{(n-1)n}{2}\right)^2

we get

∑k=1n−1(k2−k)=(n−1)n(2n−1)6−(n−1)n2=(n−1)n(n−2)3\sum_{k=1}^{n-1}(k^2-k)=\frac{(n-1)n(2n-1)}{6}-\frac{(n-1)n}{2}=\frac{(n-1)n(n-2)}{3}

and

∑k=1n−1(k3−k2)=((n−1)n2)2−(n−1)n(2n−1)6\sum_{k=1}^{n-1}(k^3-k^2)=\left(\frac{(n-1)n}{2}\right)^2-\frac{(n-1)n(2n-1)}{6}

After simplification, the numerator becomes

N=n(n−1)(n−2)(n+1)12N=\frac{n(n-1)(n-2)(n+1)}{12}

Now the denominator is

D=∑k=1nk3−∑k=1nk2D=\sum_{k=1}^{n}k^3-\sum_{k=1}^{n}k^2

Using

∑k=1nk3=(n(n+1)2)2,∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2, \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

we obtain

D=n(n+1)(3n2−n−2)12=n(n+1)(3n+2)(n−1)12D=\frac{n(n+1)(3n^2-n-2)}{12}=\frac{n(n+1)(3n+2)(n-1)}{12}

Therefore,

L=lim⁡n→∞n(n−1)(n−2)(n+1)12n(n+1)(3n+2)(n−1)12=lim⁡n→∞n−23n+2=13L=\lim_{n\to\infty}\frac{\frac{n(n-1)(n-2)(n+1)}{12}}{\frac{n(n+1)(3n+2)(n-1)}{12}}=\lim_{n\to\infty}\frac{n-2}{3n+2}=\frac{1}{3}

Therefore, the correct option is B.

Common mistakes

  • Expanding the numerator incorrectly by missing the factor n−kn-k. First rewrite it as a summation carefully, then separate the sums term by term.

  • Using the formulas for ∑k2\sum k^2 and ∑k3\sum k^3 with the wrong upper limit. The numerator runs only up to n−1n-1, so substitute n−1n-1 there before simplifying.

  • Taking the highest-power terms too early without first obtaining the exact polynomial forms. Simplify both numerator and denominator into factored expressions before evaluating the limit.

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