Question

Find the first three terms in the expansion of :
$\frac{1}{(1+x{)}^2\sqrt{1+4x}}$

Answer 1

$(1+x{)}^n=1+nx+(n\left)\right(n-1)\frac{x^2}{2!}+......(n!)\frac{x^n}{n!}$

Where n is negative or fraction

let
$y=(1+x{)}^{-2}(1+4x{)}^{-\frac{1}{2}}$
$=(1-2x+6.\frac{x^2}{2!}........)(1-2x+\frac{3}{4}.\frac{16x^2}{2!}..........)$
$=1-4x+13x^2.......$