<s>[INST] Expansion of $\frac{1}{\sqrt{1-4x}}$ Expand $\frac{1}{\sqrt{1-4x}}$ in ascending of power of $x$, up to and including the term in $x^2$, simplifying the coefficient. Hence find the coefficient of $x^2$ in the expansion of $\frac{1+2x}{\sqrt{4-16x}}$ 
My attempt, $(1-4x)^{-\frac{1}{2}}$
$=1+(-\frac{1}{2})(-4x)+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}(-4x)^2$
$=1+2x+6x^2$
How to proceed to the next part by using the result of first part?
 [/INST] $$ [x^2]\frac{1+2x}{\sqrt{4-16 x}} = \frac{1}{2}[x^2]\frac{1}{\sqrt{1-4x}}+[x]\frac{1}{\sqrt{1-4x}}\tag{1} $$
and since:
$$ \frac{1}{\sqrt{1-4x}} = 1+2x+6x^2+O(x^3)\tag{2} $$
we have:
$$ [x^2]\frac{1+2x}{\sqrt{4-16 x}} = \frac{1}{2}\cdot 6+2=\color{red}{5}.\tag{3}$$
</s>