题目详情 - Q20260126215748291

由$3a_{n+1}+a_n=\left(-\frac{1}{3}\right)^n$,得$3^{n+1}a_{n+1}+3^na_n=(-1)^n$, 若$n=2k-1,k\in N^*$, 则$\begin{cases}3^{2k}a_{2k}+3^{2k-1}a_{2k-1}=-1,\\3^{2k+1}a_{2k+1}+3^{2k}a_{2k}=1,\end{cases}$ 故$3^{2k}a_{2k}+3^{2k-1}a_{2k-1}=-1$, $3^{2k}a_{2k}+3^{2k+1}a_{2k+1}=1$, 所以$3^{2k-1}a_{2k-1}-3^{2k+1}a_{2k+1}=-2$, 解得$3^{2k+1}a_{2k+1}=1$, 故$3^{2k}a_{2k}+3^{2k-1}a_{2k-1}=-1$, 所以$3^{2k}a_{2k}+3^{2k-1}a_{2k-1}=-1$, 得$3^{2k}a_{2k}=-2$, 故$3^{2k}a_{2k}+3^{2k-1}a_{2k-1}=-1$, 所以$a_{2k}=-\frac{2}{3^{2k}}$, 又$3a_1=3,3a_2=-4$,所以$3^na_n=\begin{cases}3+2\times\frac{n-1}{2},&n=2k-1, \\ -4+(-2)\times\frac{n-2}{2},&n=2k,\end{cases}\ k\in N^*$, 即$a_n=\begin{cases}\frac{n+2}{3^n},&n=2k-1, \\ -\frac{n+2}{3^n},&n=2k,\end{cases}\ k\in N^*$, 若$n=2k-1,k\in N^*$,则$\left(\frac{n+2}{3^n}+t+\frac{n}{3^n}\right)\cdot\left(-\frac{n+3}{3^{n+1}}+t+\frac{n+1}{3^{n+1}}\right)=\left(t+\frac{2}{3^n}\right)\left(t-\frac{2}{3^{n+1}}\right)<0$ 能成立， 所以$-\frac{2}{3^n}