Find the coefficient of
$x^{6}$ in the expansion of $(1 + 2 x - x^{2})^{5}$ .

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Solution

General term of expansion is $\frac{5!}{a! b! c!} 2^{b} (- 1)^{c} x^{b + 2 c}$ , where $a + b + c = 5$

We want to find the coefficient of $x^{6}$ , therefore $b + 2 c = 6$ . This is possible for,

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

$∴$ the coefficient of $x^{6} = \frac{5!}{2! 0! 3!} 2^{0} (- 1)^{3} + \frac{5!}{1! 2! 2!} 2^{2} (- 1)^{2} + \frac{5!}{0! 4! 1!} 2^{4} (- 1)^{1} = - 10 + 120 - 80 = 30$

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