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99 Percentile Qs Bank for JEE MainMathematicsBinomial Theorem

The coefficient of x n in the polynomial ( x + 2 n + 1 C 0 ) ( x + 2 n + 1 C 1 ) ( x + 2 n + 1 C 2 ) ..... ( x + 2 n + 1 C n ) is

Options

  1. A2 n + 1
  2. B2 2 n + 1 − 1
  3. C2 2 n − 1
  4. D2 2 n

Correct answer

D. 2 2 n

Step-by-step solution

Given Expression is, ( x + 2 n + 1 C 0 ) ( x + 2 n + 1 C 1 ) ( x + 2 n + 1 C 2 ) ..... ( x + 2 n + 1 C n ) If P is coefficient of x n then, P = 2 n + 1 C 0 + 2 n + 1 C 1 + 2 n + 1 C 2 + ..... + 2 n + 1 C n -------(1) ⇒ P = 2 n + 1 C 2 n + 1 + 2 n + 1 C 2 n + 2 n + 1 C 2 n − 1 + ..... + 2 n + 1 C n + 1 ------(2) ( ∵ n C r = n C n − r ) adding (1) and (2) 2 P = ( 2 n + 1 C 0 + 2 n + 1 C 1 + ....... + 2 n + 1 C 2 n + 1 ) 2 P = 2 2 n + 1 ∴ P = 2 2 n

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