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STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
MCQ
$\frac{1}{{{{(2 + x)}^4}}} = $
  • A
    $\frac{1}{2}\left( {1 - 2x + \frac{5}{2}{x^2} - ....} \right)$
  • ✓
    $\frac{1}{{16}}\left( {1 - 2x + \frac{5}{2}{x^2} - ....} \right)$
  • C
    $\frac{1}{{16}}\left( {1 + 2x + \frac{5}{2}{x^2} + ....} \right)$
  • D
    $\frac{1}{2}\left( {1 + 2x + \frac{5}{2}{x^2} + ....} \right)$
✓

Answer

Correct option: B.
$\frac{1}{{16}}\left( {1 - 2x + \frac{5}{2}{x^2} - ....} \right)$
b
(b) ${(x + 2)^{ - 4}} = {2^{ - 4}}{\left[ {1 + \frac{x}{2}} \right]^{ - 4}} = \frac{1}{{16}}\left[ {1 - 2x + \frac{5}{2}{x^2} - ....} \right]$

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