4. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n+1}=a_{n}+4 \sqrt{a_{n}+1}+4(n \geqslant 1)$, then $a_{n}=$ $\qquad$ .
$a_{n+1}+1=\left(\sqrt{a_{n}+1}+2\right)^{2} \Rightarrow \sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+2$, and $\sqrt{a_{1}+1}=1$, so $\sqrt{a_{n+1}+1}=2 n-1 \Rightarrow a_{n}=4 n^{2}-4 n\left(n \in \mathbf{N}^{*}\right)$.