9. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$.
9. 0 .

From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$, we know
$$
S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4} \text {. }
$$

Then $a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{n-1}+1\right)^{2}$
$$
\Rightarrow\left(a_{n}-1\right)^{2}=\left(a_{n-1}+1\right)^{2}
$$
$\Rightarrow a_{n}=a_{n-1}+2$ (discard) or $a_{n}=-a_{n-1}$.
Thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and the common ratio -1.
$$
\text { Hence } S_{20}=\frac{1 \times\left[1-(-1)^{20}\right]}{1-(-1)}=0 \text {. }
$$