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Challenging problem
Solution
Let’s start, we have
\[\begin{aligned}
a_{n+1} &=2^{n}-3 a_{n}=2^{n}-3 \times 2^{n-1}+3^{2} a_{n-1} \\
&=\cdots=2^{n}+\sum_{j=1}^{n}(-3)^{j} 2^{n-j}+(-3)^{n+1} a_{0} \\
&=\frac{2^{n+1}}{5}+\left(a_{0}-\frac{1}{5}\right)(-3)^{n+1}
\end{aligned}\]
If \(a_{0}=\frac{1}{5}\), then we have no problem. If \(a_{0}\neq\frac{1}{5}\), \(\left(\frac{2}{3}\right)^{n}\) goes to 0 when \(n\rightarrow\infty\), so \(\frac{a_{n}}{3^{n}}\) will have the same sign as \(\left(a_{0}-\frac{1}{5}\right)\left(-1\right)^{n}\) when \(n\) is large. Hence, \(a_{n}<a_{n+1}\) will not hold then contradiction
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