6. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=4 a_{n}+4 \sqrt{a_{n}}+1$, $n=1,2, \cdots$. Then the general term formula of the sequence is $\qquad$ .
$$
6.2^{2 n}-2^{n+1}+1
$$

From the given condition, we have $\sqrt{a_{n+1}}=2 \sqrt{a_{n}}+1$, so,
$$
\sqrt{a_{n+1}}+1=2\left(\sqrt{a_{n}}+1\right) \text {. }
$$

Therefore, $\left\{\sqrt{a_{n+1}}+1\right\}$ is a geometric sequence with the first term 2 and the common ratio 2. $\sqrt{a_{n}}+1=2^{n}$, which means $\sqrt{a_{n}}=2^{n}-1$. Thus,
$$
a_{n}=2^{2 n}-2^{n+1}+1 \text {. }
$$