11. Given the sequence $\left\{a_{n}\right\}$, where the adjacent terms $a_{n}$ and $a_{n+1}$ are the two real roots of the equation
$$
x^{2}-n x+c_{n}=0 \quad (n=1,2, \cdots)
$$
and $a_{1}=1$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=c_{2 n-1}$, find the general term formula for the sequence $\left\{b_{n}\right\}$ and the sum of the first $n$ terms of $\left\{b_{n}\right\}$, denoted as $T_{n}$.
(If necessary, you can use:
$$
1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6} .)
$$
(1) According to the problem, by Vieta's formulas, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}=n, \quad c_{n}=a_{n} a_{n+1} . \\
\text { Then }\left(a_{n+1}+a_{n+2}\right)-\left(a_{n}+a_{n+1}\right) \\
=(n+1)-n=1,
\end{array}
$$

which implies $a_{n+2}-a_{n}=1$.
Thus, $a_{1}, a_{3}, \cdots$ and $a_{2}, a_{4}, \cdots$ are both arithmetic sequences with a common difference of 1.
Noting that $a_{1}=1, a_{2}=1-a_{1}=0$.
Therefore, for any $k \in \mathbf{Z}_{+}$, we have
$$
a_{2 k-1}=k, \quad a_{2 k}=k-1 \text {, }
$$

which means $a_{n}=\left\{\begin{array}{ll}\frac{n+1}{2}, & n \text { is odd; } \\ \frac{n-2}{2}, & n \text { is even. }\end{array}\right.$
(2) From (1), we know
$$
\begin{array}{l}
b_{n}=c_{2 n-1}=a_{2 n-1} a_{2 n} \\
=n(n-1)=n^{2}-n .
\end{array}
$$

Thus, $T_{n}=\sum_{k=1}^{n}\left(k^{2}+k\right)$
$$
\begin{array}{l}
=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2} \\
=\frac{n(n-1)(n+1)}{3} .
\end{array}
$$