11. (20 points) Let the sequence $\left\{A_{n}\right\}$ satisfy
$$
\begin{array}{l}
A_{1}=1, A_{2}=3, A_{n}=4 A_{n-1}-A_{n-2} . \\
\text { Find } \sum_{n=1}^{+\infty} \operatorname{arccot} 2 A_{n}^{2} \text { . }
\end{array}
$$
11. From the given conditions, we have
$$
\begin{array}{l}
A_{3}=4 A_{2}-A_{1}=4 \times 3-1=11, \\
\frac{A_{0}}{A_{1}}=\frac{4 A_{1}-A_{2}}{A_{1}}=1 .
\end{array}
$$

Solving the characteristic equation \(x^{2}-4 x+1=0\), we get
$$
x_{1}=2+\sqrt{3}, x_{2}=2-\sqrt{3} \text {. }
$$

Thus, the general term formula is
$$
\begin{array}{l}
A_{n}=\frac{3-\sqrt{3}}{6}(2+\sqrt{3})^{n}+\frac{3+\sqrt{3}}{6}(2-\sqrt{3})^{n} \text {, } \\
\lim _{n \rightarrow+\infty} \operatorname{arccot} \frac{A_{n}}{A_{n+1}}=\operatorname{arccot} \frac{1}{2+\sqrt{3}} \\
=\operatorname{arccot}(2-\sqrt{3})=\frac{5 \pi}{12} \text {. } \\
\text { Also, } A_{n}^{2}-A_{n+1} A_{n-1}=A_{n}^{2}-\left(4 A_{n}-A_{n-1}\right) A_{n-1} \\
=A_{n-1}^{2}+A_{n}\left(A_{n}-4 A_{n-1}\right) \\
=A_{n-1}^{2}-A_{n} A_{n-2}=\cdots=A_{2}^{2}-A_{3} A_{1} \\
=9-11 \times 1=-2 \text {, } \\
2 A_{n}^{2}=\frac{4 A_{n}^{2}}{2}=\frac{A_{n}\left(A_{n-1}+A_{n+1}\right)}{A_{n-1} A_{n+1}-A_{n}^{2}} \\
=\frac{1+\frac{A_{n-1}}{A_{n}} \cdot \frac{A_{n}}{A_{n+1}}}{\frac{A_{n-1}}{A_{n}}-\frac{A_{n}}{A_{n+1}}}, \\
\end{array}
$$

Thus, \(\operatorname{arccot} 2 A_{n}^{2}=\operatorname{arccot} \frac{A_{n}}{A_{n+1}}-\operatorname{arccot} \frac{A_{n-1}}{A_{n}}\).
Therefore, \(\sum_{n=1}^{+\infty} \operatorname{arccot} 2 A_{n}^{2}\)
$$
\begin{array}{l}
=\lim _{n \rightarrow+\infty} \operatorname{arccot} \frac{A_{n}}{A_{n+1}}-\operatorname{arccot} \frac{4 A_{1}-A_{2}}{A_{1}} \\
=\frac{5 \pi}{12}-\operatorname{arccot} 1=\frac{5 \pi}{12}-\frac{\pi}{4}=\frac{\pi}{6} .
\end{array}
$$