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KCET2014MathematicsProbability

If ( 21^ st ) and ( 22^ nd ) terms in the expansion of ( (1+x)⁴⁴ ) are equal, then ( x ) is equal to

Options

  1. A( 21 22 )
  2. B( 23 24 )
  3. C( 8 7 )
  4. D( 7 8 )

Correct answer

D. ( 7 8 )

Step-by-step solution

Given that ( (1+x)⁴⁴ ) [ array l Now, 44 20 x²⁰= 44 21 x²¹ x= 44 20 44 2 = (44-21) ! 21 ! (44-20) ! 20 ! = 23 ! 24 ! 21 ! 20 ! = 21 24 = 7 8 array ]

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