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JEE Main201910 Apr 2019Evening ShiftMathematicsBinomial TheoremActual

The smallest natural number n , such that the coefficient of x in the expansion of x 2 + 1 x 3 n is C 23 n , is

Options

  1. A58
  2. B38
  3. C35
  4. D23

Correct answer

B. 38

Step-by-step solution

In the expansion of x 2 + 1 x 3 n the general term is T r + 1 = C r   n x 2 n - r 1 x 3 r = C r   n x 2 n - 2 r - 3 r = C r   n x 2 n - 5 r For coefficient of x , 2 n - 5 r = 1 ⇒ r = 2 n - 1 5 So, we have the coefficient as C 2 n - 1 5   n Using, the given value and C r n = C n - r n ⇒ C 2 n - 1 5   n = C 23 = C n - 23   n   n If 2 n - 1 5 = 23 ⇒ n = 58 and if 2 n - 1 5 = n - 23 ⇒ n = 38 Thus, the minimum value of ' n ' is 38 .

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