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

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Solution

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

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

We want to find the coefficient of $x^{4}$ , therefore $b + 2 c = 4$ . This is possible for,

$p = 2, b = 0, c = 2; p = 3, b = 2, c = 1; p = 4, b = 4, c = 0$

$∴$ the coefficient of $x^{4} = \frac{(- 2) (- 3)}{2!} {(\frac{3}{2})}^{2} + \frac{(- 2) (- 3) (- 4)}{2!} (- 2)^{2} (\frac{3}{2}) + \frac{(- 2) (- 3) (- 4) (- 5)}{4!} (- 2)^{4} = \frac{27}{4} - 72 + 80 = \frac{59}{4}$

Therefore the required coefficient $= \frac{59}{16}$

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