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JEE Main 20236 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 6 April · Morning Shift · Mathematics · Binomial Theorem

If C3 2n : C3 n = 10: 1, then the ratio ( n ^ 2 + 3n): ( n ^ 2 - 3n...

If C32n:C3n=10:1,C3 2n : C3 n = 10: 1, then the ratio \left({n}^{2} + 3n\right): \left({n}^{2} - 3n + 4\right) is

Options

  1. A35:1635: 16
  2. B27:1127: 11
  3. C65:3765: 37
  4. D2:12: 1

Correct answer

D. 2:12: 1

Step-by-step solution

We have been given that C32n:C3n=10C3 2n : C3 n = 10

We know that Crn=n!r!(n−r)!Cr n = \dfrac{n !}{r ! \left(n - r\right) !}

⇒(2n)!3!(2n−3)!n!3!(n−3)!=10⇒ \dfrac{\dfrac{\left(2n\right) !}{3 ! \left(2n - 3\right) !}}{\dfrac{n !}{3 ! \left(n - 3\right) !}} = 10

⇒(2n)!3!(2n−3)!×3!(n−3)!n!=10⇒ \dfrac{\left(2n\right) !}{3 ! \left(2n - 3\right) !} \times \dfrac{3 ! \left(n - 3\right) !}{n !} = 10

⇒(2n)(2n−1)(2n−2)(2n−3)!(2n−3)!×(n−3)!n(n−1)(n−2)(n−3)!=10⇒ \dfrac{\left(2n\right)\left(2n - 1\right)\left(2n - 2\right)\left(2n - 3\right) !}{\left(2n - 3\right) !} \times \dfrac{\left(n - 3\right) !}{n\left(n - 1\right)\left(n - 2\right)\left(n - 3\right) !} = 10

⇒(2n)(2n−1)(2n−2)n(n−1)(n−2)=10⇒ \dfrac{\left(2n\right)\left(2n - 1\right)\left(2n - 2\right)}{n\left(n - 1\right)\left(n - 2\right)} = 10

⇒(4)(2n−1)(n−2)=10⇒ \dfrac{\left(4\right)\left(2n - 1\right)}{\left(n - 2\right)} = 10

⇒8n−4=10n−20⇒ 8n - 4 = 10n - 20

⇒n=8⇒ n = 8

Hence, the value of n2+3nn2−3n+4\dfrac{{n}^{2} + 3n}{{n}^{2} - 3n + 4} is =82+3(8)82−3(8)+4=8844=2= \dfrac{{8}^{2} + 3\left(8\right)}{{8}^{2} - 3\left(8\right) + 4} = \dfrac{88}{44} = 2

Therefore, the value of n2+3nn2−3n+4\dfrac{{n}^{2} + 3n}{{n}^{2} - 3n + 4} is 2:12: 1.

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