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JEE Mathematics 2023 Question with Solution

lim⁡n→∞(3n[4+(2+1n)2+(2+2n)2+⋯+(3−1n)2])\lim_{n \to \infty} \left( 3n \left[ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right] \right)

  • A

    1212

  • B

    193\frac{19}{3}

  • C

    00

  • D

    1919

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

lim⁡n→∞(13n∑r=0n−1(2+rn)2)\lim_{n \to \infty} \left( \frac{1}{3n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2 \right)

Find: The value of the limit.

From the given expression, write it in the Riemann sum form shown in the solution:

lim⁡n→∞13n∑r=0n−1(2+rn)2\lim_{n \to \infty} \frac{1}{3n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2

Using the standard interpretation of a Riemann sum,

=3∫01(2+x)2 dx= 3 \int_0^1 (2+x)^2 \, dx

Now evaluate the integral:

3∫01(2+x)2 dx=27−8=193 \int_0^1 (2+x)^2 \, dx = 27 - 8 = 19

Therefore, the correct option is D, and the value of the limit is 1919.

Riemann Sum Identification

Given: The terms inside the bracket are

4,(2+1n)2,(2+2n)2,…,(3−1n)24, \left(2+\frac{1}{n}\right)^2, \left(2+\frac{2}{n}\right)^2, \dots, \left(3-\frac{1}{n}\right)^2

which match

(2+rn)2forr=0,1,2,…,n−1\left(2+\frac{r}{n}\right)^2 \quad \text{for} \quad r=0,1,2,\dots,n-1

Find: Convert the sum into a definite integral.

Here the subinterval width is

Δx=1n\Delta x = \frac{1}{n}

and the sample point is

xr=rnx_r = \frac{r}{n}

so the expression corresponds to a Riemann sum for

(2+x)2(2+x)^2

on the interval

[0,1][0,1]

Thus the limit is evaluated as the integral stated in the solution:

3∫01(2+x)2 dx3 \int_0^1 (2+x)^2 \, dx

which gives

1919

Hence, the correct answer is 1919, so the correct option is D.

Common mistakes

  • Mistake: Treating the expression as a direct algebraic limit term-by-term instead of a Riemann sum. Why it is wrong: the sequence is built from many terms depending on nn, so termwise limiting misses the sum structure. Do instead: identify the pattern (2+rn)2\left(2+\frac{r}{n}\right)^2 with Δx=1n\Delta x = \frac{1}{n} and convert it to an integral.

  • Mistake: Using incorrect bounds for the integral, such as [2,3][2,3] instead of [0,1][0,1]. Why it is wrong: the variable part is x=rnx=\frac{r}{n}, so xx runs from 00 to 11. Do instead: write the integrand as f(2+x)f(2+x) with x∈[0,1]x \in [0,1].

  • Mistake: Missing the constant factor outside the Riemann sum. Why it is wrong: the final value changes if the multiplier is ignored or inverted. Do instead: carefully preserve the prefactor exactly as converted in the solution before evaluating the integral.

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