(9) Let $\left\{a_{n}\right\}$ 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 Hint: 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 $S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4}$, so
$$
a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{n-1}+1\right)^{2},
$$

which means $\left(a_{n}-1\right)^{2}=\left(a_{n-1}+1\right)^{2}$, thus $a_{n}=a_{n-1}+2$ (discard), or $a_{n}=$ $-a_{n-1}$. Therefore, $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and the common ratio -1, so $S_{20}=$ $\frac{1 \times\left[\left(1-(-1)^{20}\right]\right.}{1-(-1)}=0$.