7. In the sequence $\left\{a_{n}\right\}$, if $a_{1}=3, a_{n+1}=2 a_{n}-1\left(n \geqslant 1, n \in \mathbf{N}_{+}\right)$, then $a_{n}=$
7. $a_{n}=2^{n}+1$.

From $a_{n+1}=2 a_{n}-1$ we get $\left(a_{n+1}-1\right)=2\left(a_{n}-1\right)$, so $a_{n}-1=\left(a_{1}-1\right) 2^{n-1}$, which means $a_{n}=2^{n}+1$.