Find the coefficient of $x^{3}$ in the expansion of $(1 - 2 x + 3 x^{2} - 4 x^{3})^{\frac{1}{2}}$ .

1

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- 1

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0

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None of the above

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Solution

The correct option is B $- 1$
We know that the binomial expansion of ${(1 + x)}^{0.5} = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{16} x^{3} - \frac{5}{128} x^{4} + . . . .$
Now replace $x$ with $- 2 x + 3 x^{2} - 4 x^{3}$ in the above expansion
We get ${(1 - 2 x + 3 x^{2} - 4 x^{3})}^{0.5} = 1 + \frac{1}{2} (- 2 x + 3 x^{2} - 4 x^{3}) - \frac{1}{8} {(- 2 x + 3 x^{2} - 4 x^{3})}^{2} + \frac{1}{16} {(- 2 x + 3 x^{2} - 4 x^{3})}^{3} + . . . .$
Since we need coefficient of $x^{3}$ , consider only first four terms
The coefficient of $x^{3}$ is $- 2 + \frac{3}{2} - \frac{1}{2} = - 1$

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x^{3}

(1 - 2 x + 3 x^{2} - 4 x^{3})^{\frac{1}{2}}

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(1 - 3 x + 3 x^{2})^{- \frac{1}{2}}

The coefficient of $x^{n}$ in the expansion of $(1 + 2 x + 3 x^{2} + . . .)^{1 / 2}$ is equal to:

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will be

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