Question

"(b) use part (a) to find a power series for f(x) = 1 (4 + x)3 ."

Answer 1

Assuming

$f(x)=\dfrac1{(4+x)^3}$

Recall that for $|x|$,

$\displaystyle\sum_{n\ge0}x^n=\frac1{1-x}$

Denote the above by $s(x)$. Then

$s'(x)=\displaystyle\sum_{n\ge1}nx^{n-1}=\frac1{(1-x)^2}$
$s''(x)=\displaystyle\sum_{n\ge2}n(n-1)x^{n-2}=\frac2{(1-x)^3}$

Now,

$\dfrac1{(4+x)^3}=\dfrac1{4^3}\dfrac1{\left(1-\left(-\frac x4\right)\right)^3}=\dfrac1{2^7}\dfrac2{\left(1-\left(-\frac x4\right)\right)^3}$

which means we have

$f(x)=\dfrac1{2^7}s''(x)=\dfrac1{128}\displaystyle\sum_{n\ge2}n(n-1)\left(-\frac x4\right)^{n-2}$

which is valid for $\left|-\dfrac x4\right|$, or $|x|$.