<s>[INST] Binomial Expansion of (1+4x) I have been asked to prove the following: 
$a(n) = \frac{1}{n+1}\ \binom{2n}{n}$, given that,
$f(x) = \sum_{n = 0}^{\infty} a(n) x^n$ and $x\{f(x)\}^2$ = $f(x) -1.$
So far I have divided $x\{f(x)\}^2 = f(x) -1$ through by $x$ so I have a quadratic to work with, $\{f(x)\}^2 - \frac{f(x)}{x} -\frac{1}{x}$. 
from there I have  rearranged the equation to,
$\left(f(x)^2 -\frac{1}{2x}\right)^2 = \frac{1}{4x^2} - \frac{1}{x}$ and therefore $f(x) = \frac{1}{2x} \pm \sqrt{\frac{1}{4x^2}- \frac{1}{x} }$.
I then took $\frac{1}{2x^2}$ out as a common factor to give me $\frac{1}{2x^2}\left(1 \pm \sqrt {\frac{1}{2x} -2 }\right) $.
I then took the square root and have rearranged that to $(1-4x)^{1/2}$ which I have expanded to $$ 1 + \sum_{k=1}^{\infty} \frac{-2(2k-2)!}{k!(k-1)!} x^k.$$ 
I have no idea if this is correct or if I'm even going in the right direction but any help at all is greatly appreciated. 
 [/INST] Notice that $f(x) = 1 +  \sum_{n=1}^{\infty} a_nx^n$, so that
$$\frac{f(x)-1}{x} = \sum_{n=1}^{\infty} a_nx^{n-1} = \sum_{n=0}^{\infty} a_{n+1}x^{n}$$
With this in mind, it suffices to show that writing $f(x)^2 = \sum_{n=0}^{\infty}b_nx^n$ we have $b_n = a_{n+1}$.
Do you think you can take it from here?
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