100 and More Conjectures from the OEIS/Proof 1

First, we put the expression $(1-4x)^{3/2}\,$ into the form $(1-4x)\sqrt{1-4x} = \sqrt{1-4x}-4x{\sqrt{1-4x}}$. Now, using the binomial theorem we obtain

$(1-4x)^{\,1/2}=1+\frac12(-4x)-\frac14\frac{(-4x)^2}{2}+ \frac38\frac{(-4x)^3}{3!}-\frac{15}{16}\frac{(-4x)^4}{4!}+\cdots=\sum_{n\ge0}^{\infty}\tfrac{\tbinom{2n}{n}}{1-2n}x^n$

Likewise, $[x_n]\,4x\sqrt{1-4x} = 4\tfrac{\tbinom{2n-2}{n-1}}{3-2n}$, so we just need to manipulate factorials to get the desired result:

$\begin{align} \,[x_n]\,(1-4x)^{3/2} & = \frac{\tbinom{2n}{n}}{1-2n} - 4\frac{\tbinom{2n-2}{n-1}}{3-2n} \\ & = \frac{4n^2(2n-1)(2n-2)! - (2n-3)(2n)!}{(2n-1)(2n-3)n!^2} \\ & = \frac{4n^2(2n-1)(2n-2)(2n-4)! - (2n)!}{(2n-1)n!^2} \\ & = \frac{4n^2(2n-2)(2n-4)! - 2n(2n-2)!}{n!^2} \\ & = \frac{(2n-4)![4n^2(2n-2)-2n(2n-2)(2n-3)]}{n!n(n-1)(n-2)!} \end{align}$