<math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>n</mi> <mo>→</mo> <mo>∞</mo> </mrow> </munder> <mstyle displaystyle="true"> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> </mrow> </munderover> <mrow> <mfrac> <mrow> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>3</mn> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </mstyle> </mrow> </math>

l i m n → ∞ ∑ k = 1 n n 3 n 4 ( 1 + k 2 n 2 ) ( 1 + 3 k 2 n 2 ) = l i m n → ∞ 1 n ∑ k − 1 n 1 ( 1 + k 2 n 2 ) ( 1 + 3 k 2 n 2 ) = ∫ 0 1 d x 3 ( 1 + x 2 ) ( 1 3 + x 2 ) = ∫ 0 1 1 3 × 3 2 ( x 2 + 1 ) − ( x 2 + 1 3 ) ( 1 + x 2 ) ( x 2 + 1 3 ) d x = 1 2 [ 3 t a n − 1 ( 3 x ) ] 0 1 − 1 2 ( t a n − 1 x ) 0 1 = 3 2 ( π 3 ) − 1 2 ( π 4 ) = π 2 3 − π 8

<math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>n</mi> <mo>→</mo> <mo>∞</mo> </mrow> </munder> <mstyle displaystyle="true"> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> </mrow> </munderover> <mrow> <mfrac> <mrow> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>3</mn> <msup> <mrow> <mi>k</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> </mrow> </mstyle> </mrow> </math>

Ask & Answer Question

Limits and Derivatives Ncert Solutions Maths class 11th Maths Class 11th

$\underset{n \to \infty}{l i m} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})}$

Option 1 -

$\frac{π}{2 \sqrt[]{3}} - \frac{π}{8}$

Option 2 -

$\frac{π}{2 \sqrt[]{3}} + \frac{π}{8}$

Option 3 -

$\frac{π}{2} - \frac{π}{\sqrt[]{3}}$

Option 4 -

$\frac{π}{\sqrt[]{3}} - \frac{π}{4}$

3 Views | Posted 6 days ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

6 days ago

Correct Option - 1

Detailed Solution:

$\underset{n \to \infty}{l i m} \sum_{k = 1}^{n} \frac{n^{3}}{n^{4} (1 + \frac{k^{2}}{n^{2}}) (1 + \frac{3 k^{2}}{n^{2}})}$

$= \underset{n \to \infty}{l i m} \frac{1}{n} \sum_{k - 1}^{n} \frac{1}{(1 + \frac{k^{2}}{n^{2}}) (1 + \frac{3 k^{2}}{n^{2}})}$

$= \int_{0}^{1} \frac{d x}{3 (1 + x^{2}) (\frac{1}{3} + x^{2})}$

$= \int_{0}^{1} \frac{1}{3} \times \frac{3}{2} \frac{(x^{2} + 1) - (x^{2} + \frac{1}{3})}{(1 + x^{2}) (x^{2} + \frac{1}{3})} d x$

$= \frac{1}{2} {[\sqrt[]{3} t a n^{- 1} (\sqrt[]{3} x)]}_{0}^{1} - \frac{1}{2} {(t a n^{- 1} x)}_{0}^{1}$

$= \frac{\sqrt[]{3}}{2} (\frac{π}{3}) - \frac{1}{2} (\frac{π}{4}) = \frac{π}{2 \sqrt[]{3}} - \frac{π}{8}$

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$\underset{n \to \infty}{l i m} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})}$

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$\underset{n \to \infty}{l i m} \sum_{k = 1}^{n} \frac{n^{3}}{(n^{2} + k^{2}) (n^{2} + 3 k^{2})}$