Find the coefficient of $x^{6}$ in the expansion of $(1 + 4 x^{2} + 10 x^{4} + 20 x^{6})^{- \frac{3}{4}}$ .

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Solution

$(1 + 4 x^{2} + 10 x^{4} + 20 x^{6})^{- \frac{3}{4}} \equiv (1 + 4 y + 10 y^{2} + 20 y^{3})^{- \frac{3}{4}}$

Therefore we need to find the coefficient of $y^{3}$

General term of expansion is $\frac{\frac{- 3}{4} (\frac{- 3}{4} - 1) \dots (\frac{- 3}{4} - p + 1)}{b! c! d!} (4)^{b} (10)^{c} (20)^{d} x^{b + 2 c + 3 d}$ , where $p = b + c + d$

We want to find the coefficient of $y^{3}$ , 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 $y^{3} = (- \frac{3}{4}) (20) + (- \frac{3}{4}) (- \frac{7}{4}) (4) (10) + \frac{(- \frac{3}{4}) (- \frac{7}{4}) (- \frac{11}{4})}{3!} (4)^{3}$

$= - 15 + \frac{105}{2} - \frac{77}{2} = - 1$

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