题目详情 - Q20260126104856919

$\because a_{n+1}=4a_n-2^{n+1},\ \therefore \frac{a_{n+1}}{2^{n+1}}=2\cdot\frac{a_n}{2^n}-1,\ \therefore \frac{a_{n+1}}{2^{n+1}}-1=2\left(\frac{a_n}{2^n}-1\right)$. 又$\because \frac{a_1}{2^1}-1=2$，故$\{\frac{a_n}{2^n}-1\}$是以$2$为首项，$2$为公比的等比数列， $\therefore \frac{a_n}{2^n}-1=2\cdot2^{n-1}=2^n$，则$a_n=4^n+2^n$.