Write:
\displaystyle f(x) = 2^{-3}\left(1 + \dfrac k 2\right)^{-3} = \frac 1 8 \left(1 + \dfrac k 2 x\right)^{-3}
(a)
We have \displaystyle A = \frac 1 8.
(b)
The coefficient of the x^2 term in the series expansion of \displaystyle \frac 1 8 \left(1 + \dfrac k 2 x\right)^{-3} is:
\displaystyle \frac 1 8 \times \frac {-3(-4)} 2 \left(\frac k 2\right)^2 = \frac 3 {16} k^2
and we have:
\displaystyle \frac 3 {16} k^2 = 243
So:
k^2 = 81
giving:
k = \pm 9
Since k is positive, we have k = 9.
©
The coefficient of the x term in the series expansion of (2 + 9x)^{-3} is:
\displaystyle B = \frac 1 8 \times (-3) \times \left(\frac 9 2\right) = -\dfrac {27} {16}
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