题目详情 - Q20260127110719344

因为$a_1=\frac{5}{6}$，$a_{n+1}=\frac{1}{3}a_n+\left(\frac{1}{2}\right)^{n+1}$，所以$2^{n+1}a_{n+1}=\frac{2}{3}\times2^na_n+1$， 整理得$2^{n+1}a_{n+1}-3=\frac{2}{3}(2^na_n-3)$，所以数列$\{2^na_n-3\}$是以$2a_1-3=-\frac{4}{3}$为首项，$\frac{2}{3}$为公比的等比数列，所以$2^na_n-3=-\frac{4}{3}\left(\frac{2}{3}\right)^{n-1}$，解得$a_n=\frac{3}{2^n}-\frac{2}{3^n}$． 故答案为：$\frac{3}{2^n}-\frac{2}{3^n}$．