11. If the adjacent terms $a_{n}, a_{n+1}$ in the sequence $\left\{a_{n}\right\}$ are the two real roots of the equation $x^{2}-n x+c_{n}=0(n=1,2,3, \cdots)$, and $a_{1}=1$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=c_{2 n-1}$, find the general term formula of the sequence $\left\{b_{n}\right\}$ and the sum of the first $n$ terms $T_{n}$ of $\left\{b_{n}\right\}$.
(1) $a_{n}+a_{n+1}=n \Rightarrow a_{1}=1, a_{2}=0, a_{3}=2, a_{4}=1, a_{5}=3, a_{6}=2, a_{7}=4, a_{8}=3$, $a_{9}=5, a_{10}=4, \cdots$, conjecture $a_{n}=\left\{\begin{array}{l}\frac{n+1}{2}, n \text { is odd. } \\ \frac{n}{2}-1, n \text { is even. }\end{array}\right.$

Assume the conclusion holds for $n=k$, when $k$ is odd, $a_{k}+a_{k+1}=k \Rightarrow a_{k+1}=k-a_{k}$ $=k-\frac{k+1}{2}=\frac{k-1}{2}=\frac{k+1}{2}-1$; when $k$ is even, $a_{k}+a_{k+1}=k \Rightarrow a_{k+1}=k-a_{k}$ $=k-\frac{k}{2}+1=\frac{k}{2}+1=\frac{k+1+1}{2}$, that is, the conclusion also holds for $n=k+1$.

In summary, we have $a_{n}=\left\{\begin{array}{l}\frac{n+1}{2}, n \text { is odd. } \\ \frac{n}{2}-1, n \text { is even. }\end{array}\right.$
(2) $b_{n}=a_{2 n-1} a_{2 n}=n(n-1)=n^{2}-n\left(n \in \mathbf{N}_{+}\right)$.

Therefore, $T_{n}=\sum_{k=1}^{n} k^{2}-\sum_{k=1}^{n} k=\frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}=\frac{n\left(n^{2}-1\right)}{3}$.