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JEE Mains · Maths · STD 11 - 7. binomial theoram

\(n\) ની ન્યૂનતમ કિમત મેળવો કે જેથી \({\left( {{x^2}\, + \,\frac{1}{{{x^3}}}} \right)^n}\) ના વિસ્તરણમાં \(x\) નો સહગુણક \(^n{C_{23}}\) થાય ? 

  1. A \(38\)
  2. B \(58\)
  3. C \(23\)
  4. D \(35\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(38\)

Step-by-step Solution

Detailed explanation

\({T_{r + 1}}\, = \,\sum\limits_{r = 0}^n {^n{C_r}\,{x^{2n - 2r}}\,.\,{x^{ - 3r}}} \) \(2n - 5r = 1\Rightarrow 2n = 5r + 1\) for \(r = 15, n = 38\) smallest value of \(n\) is \(38\).
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