4. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{0}=0, x_{n+1}=x_{n}+a+\sqrt{b^{2}+4 a x_{n}}, n=0$, $1,2, \cdots$, where $a, b$ are given positive real numbers, find the general term of this sequence.
4. Let $y_{n}=\sqrt{b^{2}+4 a x_{n}}$, then $y_{0}=b, x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a}$, thus we have $y_{n+1}^{2}=\left(y_{n}+2 a\right)^{2}$. Since $y_{n}>0$, it follows that $y_{n+1}-y_{n}=2 a$, which means $\left\{y_{n}\right\}$ is an arithmetic sequence with a common difference of $2 a$. Therefore, $y_{n}=y_{0} + 2 a n = 2 a n + b$. Hence, $x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a}=a n^{2}+b n$.