Try to determine the real number $a_{0}$, such that the sequence $\left\{a_{n}\right\}$ defined by the recurrence formula $a_{n+1}=-3 a_{n}+2^{n}(n=0,1, \cdots)$ is strictly increasing, i.e., for $n \geqslant 0$ we have $a_{n}<a_{n+1}$.


15. It is known that $a_{1}=-3 a_{0}+1, a_{2}=3^{2} a_{0}-3+2, a_{3}=-3^{3} a_{0}+3^{3}-3 \times 2+2^{2}, a_{4}=3^{4} a_{0}-3^{2}+3^{2} \times 2-3 \times 2^{2}+2^{3}$, from which we can expect that in general, $a_{n}=(-3)^{n} a_{0}+(-1)^{n-1}\left(3^{n 1}-3^{n-2} \times 2+\cdots \cdots 3 \times 2^{n 2}\right)+2^{n-1}$. This can be easily proven by induction (omitted here). The above formula can be rewritten as:
$$
a_{n}=\left(a_{0}-\frac{1}{5}\right)(-3)^{n}+\frac{2^{n}}{5},
$$
which can be further transformed to:
$$
a_{n}=(-3)^{n}\left[\left(a_{0}-\frac{1}{5}\right)+\left(-\frac{2}{3}\right)^{n} \times \frac{1}{5}\right]
$$

Note: When $n$ is sufficiently large, $\left(\frac{2}{3}\right)^{n}$ tends to 0. Therefore, if $a_{0}-\frac{1}{5} \neq 0$, the term in the brackets will be close to $a_{0}-\frac{1}{5}$. Since $(-3)^{n} (n=0,1,2, \cdots)$ alternates between positive and negative values, the sequence $\{a_{n}\}$ cannot be strictly increasing. Therefore, $a_{0}-\frac{1}{5}=0$, i.e., $a_{n}=\frac{1}{5}$. In this case, $a_{n}=\frac{2^{n}}{5}$, which is strictly increasing.

The general term formula for $a_{n}$ can also be derived as follows: Rewrite the recurrence relation $a_{n+1}=-3 a_{n}+2^{n}$ as $\left(a_{n+1}-\frac{2^{n+1}}{5}\right)=-3\left(a_{n}-\frac{2^{n}}{5}\right)$. Let $b_{n}=a_{n}-\frac{2^{n}}{5}(n \geqslant 0)$, then the above equation becomes $b_{n+1}=-3 b_{n}$, so $b_{n}=(-3)^{n} b_{0}$, from which we get $a_{n}=\left(a_{0}-\frac{1}{5}\right)(-3)^{n}+\frac{2^{n}}{5}(n \geqslant 0)$.