3、Given in the sequence $\left\{a_{n}\right\}, a_{1}=1, a_{2}=3$, and for any $n \in N_{+}$, $a_{n+2} \leq a_{n}+3 \times 2^{n}$, $a_{n+1} \geq 2 a_{n}+1$ hold. Then the general term formula of the sequence $\left\{a_{n}\right\}$ is $\qquad$
3. $a_{n}=2^{n}-1$

Analysis: From $a_{n+1}+1 \geq 2\left(a_{n}+1\right)$, we get $a_{n}+1 \geq 2^{n-1}\left(a_{1}+1\right)=2^{n}$, thus $a_{n} \leq 2^{n}-1$. Furthermore, from $a_{n+2} \leq a_{n}+3 \times 2^{n}$, we know $a_{n+2}-2^{n+2} \leq a_{n}-2^{n} \leq \ldots \leq \max \left\{a_{2}-2^{2}, a_{1}-2\right\}=-1$, hence $a_{n} \leq 2^{n}-1$. Therefore, $a_{n}=2^{n}-1$.