= Solution
Write $\bar i=2n+1-i$, and let $E_{ab}$ denote a <matrix unit>. The diagonal <Cartan subalgebra> consists of
$$
t=\operatorname{diag}(t_1,\ldots,t_n,-t_n,\ldots,-t_1),\qquad \varepsilon_i(t)=t_i.
$$
For a <diagonal matrix>, $[t,E_{ab}]=(t_a-t_b)E_{ab}$. Substituting into the symplectic relation shows that the following are <root vectors>:
$$
\begin{array}{c|c}
\text{weight}&\text{vector}\\\hline
\varepsilon_i-\varepsilon_j&E_{ij}-E_{\bar j\bar i}\quad(i\ne j)\\
\varepsilon_i+\varepsilon_j&E_{i\bar j}+E_{j\bar i}\quad(i<j)\\
-\varepsilon_i-\varepsilon_j&E_{\bar i j}+E_{\bar j i}\quad(i<j)\\
2\varepsilon_i&E_{i\bar i}\\
-2\varepsilon_i&E_{\bar i i}
\end{array}
$$
The zero <weight space> is spanned by $H_i=E_{ii}-E_{\bar i\bar i}$. The symplectic equation pairs the remaining entries exactly as in this table; the displayed vectors and the $H_i$ are independent and span the algebra. This proves the <root-space decomposition>
$$
\mathfrak g=\mathfrak t\oplus\bigoplus_{\alpha\in R}\mathbb C X_\alpha,\qquad
\boxed{R=\{\pm\varepsilon_i\pm\varepsilon_j:i<j\}\cup\{\pm2\varepsilon_i:1\le i\le n\}.}
$$
There are $2n^2$ <roots of a root system>, each with one-dimensional <root space>, giving $\dim\mathfrak g=n(2n+1)$.
The upper triangular <root vectors> select precisely
$$
\boxed{R^+=\{\varepsilon_i-\varepsilon_j,\varepsilon_i+\varepsilon_j:i<j\}\cup\{2\varepsilon_i\}.}
$$
The <simple roots> of this <Cn root system> are
$$
\alpha_i=\varepsilon_i-\varepsilon_{i+1}\ (1\le i<n),\qquad \alpha_n=2\varepsilon_n.
$$
Indeed, the positive <roots of a root system> have expansions
$$
\begin{aligned}
\varepsilon_i-\varepsilon_j&=\alpha_i+\cdots+\alpha_{j-1},\\
\varepsilon_i+\varepsilon_j&=\alpha_i+\cdots+\alpha_{j-1}+2(\alpha_j+\cdots+\alpha_{n-1})+\alpha_n,\\
2\varepsilon_i&=2(\alpha_i+\cdots+\alpha_{n-1})+\alpha_n.
\end{aligned}
$$
These expansions also identify the <highest root>:
$$
\boxed{\theta=2\varepsilon_1=2\alpha_1+\cdots+2\alpha_{n-1}+\alpha_n.}
$$
Normalize the coordinate <inner product> by $(\varepsilon_i,\varepsilon_j)=\delta_{ij}$. The <simple coroots> are $\alpha_i^\vee=\varepsilon_i-\varepsilon_{i+1}$ for $i<n$ and $\alpha_n^\vee=\varepsilon_n$. Solving $\langle\omega_j,\alpha_i^\vee\rangle=\delta_{ij}$ gives the <fundamental weights>
$$
\boxed{\omega_j=\varepsilon_1+\cdots+\varepsilon_j\quad(1\le j\le n).}
$$
In the sum of the positive <roots of a root system>, the pair $(\varepsilon_i-\varepsilon_j)+(\varepsilon_i+\varepsilon_j)$ contributes $2\varepsilon_i$, and the long root contributes another $2\varepsilon_i$. Therefore the <Weyl vector> is
$$
\boxed{\rho=\frac12\sum_{\alpha\in R^+}\alpha=\sum_{i=1}^n(n-i+1)\varepsilon_i=\sum_{j=1}^n\omega_j.}
$$
These are the <Positive-root data for Cn>.
For $n\ge2$, the finite <Dynkin diagram> is a chain of $n$ vertices. The bonds between $\alpha_i$ and $\alpha_{i+1}$ for $i<n-1$ are single. The last bond is double, with its arrow pointing from the <long root> $\alpha_n$ toward the <short root> $\alpha_{n-1}$. For the <Extended Dynkin diagram>, add $\alpha_0=-\theta=-2\varepsilon_1$ at the other end, doubly bonded to $\alpha_1$, with arrow toward $\alpha_1$. The following <Cn Dynkin diagrams> include both ends and the low-rank cases.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-6-cn-diagrams.png]
{title=Finite and extended Cn Dynkin diagrams with arrows toward the short roots and separate rank-one and rank-two cases}
At $n=1$, there is just the root $\alpha_1=2\varepsilon_1$ and one finite vertex. The affine <Cartan matrix> is $\begin{pmatrix}2&-2\\-2&2\end{pmatrix}$; its two vertices have equal root length, so the double bond has no arrow.
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