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

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Solution

General term of expansion is $\frac{\frac{1}{2} (\frac{1}{2} - 1) \dots (\frac{1}{2} - p + 1)}{b! c! d!} (- 2)^{b} 3^{c} (- 4)^{d} x^{b + 2 c + 3 d}$ , where $p = b + c + d$

We want to find the coefficient of $x^{3}$ , therefore $b + 2 c + 3 d = 3$ . This is possible for,

$p = 1, b = 0, c = 0, d = 1; p = 2, b = 1, c = 1, d = 0; p = 3, b = 3, c = 0, d = 0$

$∴$ the coefficient of $x^{3} = \frac{1}{2} (- 4) + \frac{1}{2} (- \frac{1}{2}) (- 2) 3 + \frac{\frac{1}{2} (- \frac{1}{2}) (- \frac{3}{2})}{3!} (- 2)^{3}$

$= - 2 + \frac{3}{2} - \frac{1}{2} = - 1$

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