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KCET2018MathematicsBinomial Theorem

The constant term in the expansion of ( (x²- 1 x² )¹⁶ ) is

Options

  1. A( 16 8 )
  2. B( 16 7 )
  3. C( 16 9 )
  4. D( 16 10 )

Correct answer

A. ( 16 8 )

Step-by-step solution

Given equation ( (x²- 1 x² )¹⁶ ) [ array l = 16 0 (x² )¹⁶ ( -1 x² )⁰+ 16 1 (x² )¹⁵ (- 1 x² )¹+ 16 2 (x² )¹⁴ ( -1 x² )² + 16 3 (x² )¹³ (- 1 x² )³+ s+ 16 8 (x² )⁸ (- 1 x² )⁸+ . array ] Thus, the constant term is [ 16 8 x¹⁶ 1 x¹⁶ = 16 8 ]

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