= Solution
Apply the <continued-fraction algorithm>. Since $3<\sqrt{11}<4$,
$$
x_0=\sqrt{11},\qquad a_0=3.
$$
The successive complete quotients are
$$
x_1=\frac1{\sqrt{11}-3}
=\frac{\sqrt{11}+3}{2},
\qquad a_1=3,
$$
and
$$
x_2=\frac1{x_1-3}
=\sqrt{11}+3,
\qquad a_2=6.
$$
Taking one more reciprocal returns to $x_1$, so the <continued fraction of the square root of eleven> is
$$
\boxed{\sqrt{11}=[3;\overline{3,6}].}
$$
Write
$$
z_n=p_n+q_n\sqrt{11}.
$$
The first convergents are $p_0/q_0=3$ and $p_1/q_1=10/3$, whence
$$
z_1=10+3\sqrt{11}
=\frac{3+\sqrt{11}}2(3+\sqrt{11})
=\frac{3+\sqrt{11}}2z_0.
$$
The <continued fraction convergent> recurrence is
$$
z_{n+1}=a_{n+1}z_n+z_{n-1},
$$
where $a_{n+1}=3$ for even $n$ and $a_{n+1}=6$ for odd $n$.
Set
$$
\alpha=3+\sqrt{11},
\qquad
\beta=\frac{3+\sqrt{11}}2.
$$
If $n$ is even and $z_{n+1}=\beta z_n$, then
$$
z_{n+2}=6z_{n+1}+z_n
=\left(6+\frac1\beta\right)z_{n+1}
=(3+\sqrt{11})z_{n+1}
=\alpha z_{n+1},
$$
because $\beta^{-1}=\sqrt{11}-3$. If $n$ is odd and $z_{n+1}=\alpha z_n$, then
$$
z_{n+2}=3z_{n+1}+z_n
=\left(3+\frac1\alpha\right)z_{n+1}
=\frac{3+\sqrt{11}}2z_{n+1}
=\beta z_{n+1},
$$
because $\alpha^{-1}=(\sqrt{11}-3)/2$. The base case and induction prove the <alternating multiplier recurrence for convergents of the square root of eleven>:
$$
\boxed{
p_{n+1}+q_{n+1}\sqrt{11}
=\begin{cases}
(3+\sqrt{11})(p_n+q_n\sqrt{11}),&n\text{ odd},\\[2mm]
\dfrac{3+\sqrt{11}}2(p_n+q_n\sqrt{11}),&n\text{ even}.
\end{cases}}
$$
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