31. Find all real numbers $a_{0}$ such that the sequence $\left\{a_{n}\right\}$, satisfying $a_{n+1}=2^{n}-3 a_{n}$, is increasing.
31. $a_{n}=2^{n-1}-3 a_{n-1}=2^{n-1}-3 \cdot 2^{n-2}+9 a_{n-2}=\cdots=2^{n-1}-3 \cdot 2^{n-2}$ $+\cdots+(-1)^{n-1} \cdot 3^{n-1}+(-1)^{n} \cdot 3^{n} a_{0}$, thus $a_{n}=\frac{2^{n}-(-3)^{n}}{5}+(-3)^{n} \cdot$ $a_{0}$. Let $d_{n}=a_{n}-a_{n-1}=\frac{2^{n}-(-3)^{n}}{5}+(-3)^{n} a_{0}-\frac{2^{n-1}-(-3)^{n-1}}{5}-$ $(-3)^{n-1} a_{0}=\frac{2^{n-1}}{5}+\frac{4}{5} \cdot(-3)^{n-1}-4 \cdot(-3)^{n-1} a_{0}=\frac{2^{n-1}}{5}+4 \cdot$ $(-3)^{n-1}\left(\frac{1}{5}-a_{0}\right)$. If $a_{0}>\frac{1}{5}$, then for sufficiently large even $n, d_{n}<0$; if $a_{0}<\frac{1}{5}$, then for sufficiently large odd $n, d_{n}<0$. Therefore, $a_{0}=\frac{1}{5}$.