1. 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.
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 $y_{n+1}^{2}=\left(y_{n}+2 a\right)^{2}$, and since $y_{n}>0$, then $y_{n+1}-$ $y_{n}=2 a$. Therefore, $\left\{y_{n}\right\}$ is an arithmetic sequence with a common difference of $2 a$. From this, we deduce that $y_{n+1}=y_{0}+2 a n=2 a n+b$. Hence, $x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a}=$ $a n^{2}+b n$.