正项数列$\left\{a_n\right\}\mathrm{满足}{a_n}^2-\left(2n-1\right)a_n-2n=0$.

$$\begin{gathered} \left(1\right) \mathrm{求数列}\left\{a_n\right\}\mathrm{通项公式}{a}_n; \\ \left(2\right) 令b_n=\frac{1}{\left(n+1\right)a_n},\mathrm{求数列}\left\{b_n\right\}前n\mathrm{项的和}{T}_n. \end{gathered}$$