If <picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753693928php2huvK3.jpeg" alt="" width="205" height="63" loading="lazy"></picture>then L is equal to…………

r ⋅ n ? C r = n ⋅ n − 1 ? C r − 1 ∑ k = 1 1 0 ( k ⋅ 1 0 ? C k ) 2 = ∑ k = 1 1 0 ( 1 0 ⋅ 9 ? C K − 1 ) 2 = 1 0 0 ⋅ 1 8 ? C 9 22000L =  1 0 0 ⋅ 1 8 ? C 9 L = 1 8 ! = 2 1 6 ⋅ 3 8 ⋅ 5 3 ⋅ 7 2 ⋅ 1 1 1 ⋅ 1 3 ⋅ 1 7 9 + 4 + 2 + 1 9 ! = 2 7 ⋅ 3 4 ⋅ 5 1 ⋅ 7 1 6 + 2 4 + 2 + 1 18!(9!)2=22⋅5⋅11⋅13⋅17 3 + 3 + 1 = 221

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Integrals Ncert Solutions Maths class 12th Maths Class 12th

If then L is equal to…………

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alok kumar singh | Contributor-Level 10

2 months ago

$r \cdot^{n} ? C_{r} = n \cdot^{n - 1} ? C_{r - 1}$

$\sum_{k = 1}^{10} {(k \cdot^{10} ? C_{k})}^{2} = \sum_{k = 1}^{10} {(10 \cdot^{9} ? C_{K - 1})}^{2}$

= $100 \cdot^{18} ? C_{9}$

22000L = $100 \cdot^{18} ? C_{9}$

L =

$18! = 2^{16} \cdot 3^{8} \cdot 5^{3} \cdot 7^{2} \cdot 1 1^{1} \cdot 13 \cdot 17$

$9 + 4 + 2 + 1$ $9! = 2^{7} \cdot 3^{4} \cdot 5^{1} \cdot 7^{1}$

6 + 2 4 + 2 + 1 $\frac{18!}{{(9!)}^{2}} = 2^{2} \cdot 5 \cdot 11 \cdot 13 \cdot 17$

3 + 3 + 1

= 221

<picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753694062phpPXRcxw.jpeg" alt="" width="232" height="57" loading="lazy"></picture> <math> <mrow> <mi>r</mi> <msup> <mrow> <mo>⋅</mo> </mrow> <mrow> <mi>n</mi> </mrow> </msup> <mtext>?</mtext> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>r</mi> </mrow> </msub> <mo>=</mo> <mi>n</mi> <msup> <mrow> <mo>⋅</mo> </mrow> <mrow> <mi>n</mi> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mtext>?</mtext> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>r</mi> <mo>−</mo> <mn>1</mn> </mrow> </msub> </mrow> </math> <math> <mrow> <mstyle displaystyle="true"> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </munderover> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>k</mi> <msup> <mrow> <mo>⋅</mo> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </msup> <mtext>?</mtext> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>k</mi> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mstyle> <mo>=</mo> <mstyle displaystyle="true"> <munderover> <mo>∑</mo> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> </mrow> </munderover> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>1</mn> <mn>0</mn> <msup> <mrow> <mo>⋅</mo> </mrow> <mrow> <mn>9</mn> </mrow> </msup> <mtext>?</mtext> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mi>K</mi> <mo>−</mo> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mstyle> </mrow> </math> = <math> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <msup> <mrow> <mo>⋅</mo> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </msup> <mtext>?</mtext> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mn>9</mn> </mrow> </msub> </mrow> </math>  22000L =   <math> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <msup> <mrow> <mo>⋅</mo> </mrow> <mrow> <mn>1</mn> <mn>8</mn> </mrow> </msup> <mtext>?</mtext> <msub> <mrow> <mi>C</mi> </mrow> <mrow> <mn>9</mn> </mrow> </msub> </mrow> </math>  L =<picture><source srcset="" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753694133php8AVqbD.jpeg" alt="" width="184" height="61" loading="lazy"></picture> <math> <mrow> <mn>1</mn> <mn>8</mn> <mo>!</mo> <mo>=</mo> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>1</mn> <mn>6</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>8</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mn>7</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>⋅</mo> <mn>1</mn> <msup> <mrow> <mn>1</mn> </mrow> <mrow> <mn>1</mn> </mrow> </msup> <mo>⋅</mo> <mn>1</mn> <mn>3</mn> <mo>⋅</mo> <mn>1</mn> <mn>7</mn> </mrow> </math> <math> <mrow> <mn>9</mn> <mo>+</mo> <mn>4</mn> <mo>+</mo> <mn>2</mn> <mo>+</mo> <mn>1</mn> </mrow> </math> <math> <mrow> <mn>9</mn> <mo>!</mo> <mo>=</mo> <msup> <mrow> <mn>2</mn> </mrow> <mrow> <mn>7</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mn>5</mn> </mrow> <mrow> <mn>1</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mn>7</mn> </mrow> <mrow> <mn>1</mn> </mrow> </msup> </mrow> </math> 6 + 2                    4 + 2 + 1   <math><mrow><mfrac><mrow><mn>1</mn><mn>8</mn><mo>!</mo></mrow><mrow><msup><mrow><mrow><mo>(</mo><mrow><mn>9</mn><mo>!</mo></mrow><mo>)</mo></mrow></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><mn>2</mn></mrow></msup><mo>⋅</mo><mn>5</mn><mo>⋅</mo><mn>1</mn><mn>1</mn><mo>⋅</mo><mn>1</mn><mn>3</mn><mo>⋅</mo><mn>1</mn><mn>7</mn></mrow></math>            3 +                       3 + 1 = 221

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I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due to OCR quality, but it seems to compare the final result with a given form to find coefficients α, β, γ, δ. The final expression shown is:
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The problem is to evaluate the integral:
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= ∑? n * e? [-e^ (1-x)] from n to n+1
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= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

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