6. The sequence $\left\{a_{n}\right\}$ is a monotonically increasing sequence, and $a_{n}=$ $2^{n-1}-3 a_{n-1}, n=1,2, \cdots$. Then the first term $a_{0}$ is equal to $\qquad$.
6. $\frac{1}{5}$.

From the given, we have $a_{n}-\frac{2^{n}}{5}=-3\left(a_{n-1}-\frac{2^{n-1}}{5}\right)$, hence $a_{n}-\frac{2^{n}}{5}=(-3)^{n}\left(a_{0}-\frac{1}{5}\right)$.
$a_{n}-a_{n-1}=\frac{2^{n-1}}{5}\left[1+4(-1)^{n-1}\left(\frac{3}{2}\right)^{n-1}\left(1-5 a_{0}\right)\right]$.
If $1-5 a_{0}>0$, then for sufficiently large even $n, a_{n}-a_{n-1}>0$