8. 173 Find all $a_{0} \in R$, such that the sequence
$$a_{n+1}=2^{n}-3 a_{n}, n=0,1,2, \cdots$$

determined by this is increasing.
[Solution] Since for $n \geqslant 0$ we have
$$\begin{aligned}
a_{n+1} & =2^{n}-3 a_{n}=2^{n}-3 \cdot 2^{n-1}+3^{2} a_{n-1} \\
& =2^{n}-3 \cdot 2^{n-1}+3^{2} \cdot 2^{n-2}-3^{3} a_{n-2} \\
& =\cdots=\sum_{k=0}^{n}(-1)^{k} 2^{n-k} \cdot 3^{k}+(-1)^{n+1} 3^{n+1} a_{0} \\
& =\frac{1}{5}\left(2^{n+1}-(-1)^{n+1} 3^{n+1}\right)+(-1)^{n+1} 3^{n+1} a_{0}
\end{aligned}$$

Therefore, when $n \geqslant 1$,
$$d_{n}=a_{n+1}-a_{n}=\frac{1}{5} \cdot 2^{n}+(-1)^{n+1} 4 \cdot 3^{n}\left(a_{0}-\frac{1}{5}\right) .$$

Clearly, $d_{0}=a_{1}-a_{0}=1-4 a_{0}$ also satisfies (1). From (1), we know that when $a_{0} \neq \frac{1}{5}$, there exists a sufficiently large $n$ such that $d_{n}>0$, and there also exists a sufficiently large $n$ such that $d_{n}<0$, which means the sequence $\left\{a_{n}\right\}$ is not monotonic.

On the other hand, when $a_{0}=\frac{1}{5}$, we have $d_{n}=\frac{1}{5} \cdot 2^{n}>0$ for all $n \geqslant 0$, which means the sequence $\left\{a_{n}\right\}$ is monotonically increasing.

In summary, $a_{0}=\frac{1}{5}$ is the required value.