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JEE Main 202330 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
30 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2023 · 30 January · Evening Shift · Mathematics · Definite Integration

_ n → ∞ (3)/(n) \ 4 + (2 + (1)/(n)) ^ 2 + (2 + (2)/(n)) ^ 2 + + (3...

lim⁡n→∞3n\lim_{n \to \infty}\dfrac{3}{n}\left\{4 + {\left(2 + \dfrac{1}{n}\right)}^{2} + {\left(2 + \dfrac{2}{n}\right)}^{2} + \ldots + {\left(3 - \dfrac{1}{n}\right)}^{2}\right\} is equal to

Options

  1. A1212
  2. B193\dfrac{19}{3}
  3. C00
  4. D1919

Correct answer

D. 1919

Step-by-step solution

Let

L=lim⁡n→∞3n[4+(2+1n)2+(2+2n)2+...+(3−1n)2]L = \lim_{n \to \infty}\dfrac{3}{n}\left[4 + {\left(2 + \dfrac{1}{n}\right)}^{2} + {\left(2 + \dfrac{2}{n}\right)}^{2} + . . . + {\left(3 - \dfrac{1}{n}\right)}^{2}\right]

⇒L=lim⁡n→∞3n[(2+0n)2+(2+1n)2+(2+2n)2+...+(2+(1−1n))2]⇒ L = \lim_{n \to \infty}\dfrac{3}{n}\left[{\left(2 + \dfrac{0}{n}\right)}^{2} + {\left(2 + \dfrac{1}{n}\right)}^{2} + {\left(2 + \dfrac{2}{n}\right)}^{2} + . . . + {\left(2 + \left(1 - \dfrac{1}{n}\right)\right)}^{2}\right]

⇒L=3lim⁡n→∞1n[∑r=0n−1(2+rn)2]⇒ L = 3\lim_{n \to \infty}\dfrac{1}{n}\left[\mathop{∑}_{r = 0}^{n - 1}{\left(2 + \dfrac{r}{n}\right)}^{2}\right]

⇒L=3∫01(2+x)2dx⇒ L = 3{∫}_{0}^{1}{\left(2 + x\right)}^{2}dx

⇒L=[(2+x)3]01⇒ L = {\left[{\left(2 + x\right)}^{3}\right]}_{0}^{1}

⇒L=27−8⇒ L = 27 - 8

⇒L=19⇒ L = 19

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