Assuming
[tex]f(x)=\dfrac1{(4+x)^3}[/tex]
Recall that for [tex]|x|<1[/tex],
[tex]\displaystyle\sum_{n\ge0}x^n=\frac1{1-x}[/tex]
Denote the above by [tex]s(x)[/tex]. Then
[tex]s'(x)=\displaystyle\sum_{n\ge1}nx^{n-1}=\frac1{(1-x)^2}[/tex]
[tex]s''(x)=\displaystyle\sum_{n\ge2}n(n-1)x^{n-2}=\frac2{(1-x)^3}[/tex]
Now,
[tex]\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}[/tex]
which means we have
[tex]f(x)=\dfrac1{2^7}s''(x)=\dfrac1{128}\displaystyle\sum_{n\ge2}n(n-1)\left(-\frac x4\right)^{n-2}[/tex]
which is valid for [tex]\left|-\dfrac x4\right|<1[/tex], or [tex]|x|<4[/tex].
