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(a) Find the first 3 terms in ascending powers of $x$ of the binomial expansion of $\left(2-\frac{x}{8}\right)^{7}$ $\mathrm{f}(x)=(a x+b)\left(2-\frac{x}{8}\right)^{7}$ where $a$ and $b$ are constants
Given that the first two terms, in ascending powers of $x$, in the series expansion of $\mathrm{f}(x)$ are 384 and $-104 x$

(b) Find the values of $a$ and $b$
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(a)

$\left(2-\frac{x}{8}\right)^{7}$
$1(2)^{7}+7(2)^{6}\left(-\frac{x}{3}\right)+21(2)^{5}\left(-\frac{x}{8}\right)^{2}$
$128-56 x+\frac{21}{2} x^{2}$

(b)

$(a x+b)\left(128-56 x+\frac{21}{2} x^{2}\right)$
$128 a x+128 b-56 a x^{2}-56 b x \ldots$
$128 b+(128 a-56 b) x \ldots$
$128 b=384 \quad 128 a-56(3)=-104$
$b=3$
$128 a=64$
$a=\frac{1}{2}$
by Platinum (122,714 points)

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