2. (1994 Bulgarian Mathematical Olympiad) The sequence of real numbers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ is defined by the following equation: $a_{n+1}=2^{n}-3 a_{n}, n=0,1,2, \cdots$
(1) Find the expression for $a_{n}$ in terms of $a_{0}$ and $n$.
(2) Find $a_{0}$ such that for any positive integer $n, a_{n+1}>a_{n}$.
2. From the given condition, we have $\frac{a_{n+1}}{2^{n+1}}=\frac{1}{2}-\frac{3}{2} \cdot \frac{a^{n}}{2^{n}}$. Let $b_{n}=\frac{a^{n}}{2^{n}}$, then $b_{n+1}=\frac{1}{2}-\frac{3}{2} b_{n}$, which means $b_{n+1}-\frac{1}{5}=-\frac{3}{2}\left(b_{n}-\right.$ $\frac{1}{5}$ ) (here $\frac{1}{5}$ is the root of the equation $x=\frac{1}{2}-\frac{3}{2} x$). Thus, $b_{n}-\frac{1}{5}=\left(b_{0}-\frac{1}{5}\right)\left(-\frac{3}{2}\right)^{n}=\left(a_{0}-\frac{1}{5}\right)\left(-\frac{3}{2}\right)^{n}$, so $a_{n}=2^{n} b_{n}=\left(a_{0}-\frac{1}{5}\right)(-3)^{n}+\frac{1}{5} \cdot 2^{n}$, which means $\frac{1}{3^{n}} a_{n}=\left(a_{0}-\frac{1}{5}\right)(-1)^{n}+\frac{1}{5}\left(\frac{2}{3}\right)^{n}$. If $a_{0}-\frac{1}{5} \neq 0$, then for sufficiently large $n$, $a_{n}$ has the same sign as $(-1)^{n}\left(a_{0}-\frac{1}{5}\right)$, making it impossible for $a_{n+1}>a_{n}$ for all $n \in \mathbf{N}_{+}$; when $a_{0}=\frac{1}{5}$, $a_{n}=\frac{1}{5} \cdot 2^{n}$, clearly $a_{n+1}>a_{n}$ for all $n \in \mathbf{N}_{+}$, hence the required $a_{0}=\frac{1}{5}$.