Question

5．已知數(shù)列{a_n}滿足${a_1}+3{a_2}+{3^2}{a_3}+…+{3^{n-1}}{a_n}=\frac{n+1}{3}$，a_n=$\left\{\begin{array}{l}{\frac{2}{3}，n=1}\\{\frac{1}{{3}^{n}}，n≥2}\end{array}\right.$．

Answer 1

分析 由已知數(shù)列遞推式可得n≥2時(shí)，${a}_{1}+3{a}_{2}+{3}^{2}{a}_{3}+…+{3}^{n-2}{a}_{n-1}=\frac{n}{3}$，與原遞推式作差可得${a}_{n}=\frac{1}{{3}^{n}}$（n≥2）．再由原遞推式求出首項(xiàng)，驗(yàn)證后得答案．

解答 解：由${a_1}+3{a_2}+{3^2}{a_3}+…+{3^{n-1}}{a_n}=\frac{n+1}{3}$，①
得n≥2時(shí)，${a}_{1}+3{a}_{2}+{3}^{2}{a}_{3}+…+{3}^{n-2}{a}_{n-1}=\frac{n}{3}$，②
①-②得：${3}^{n-1}{a}_{n}=\frac{n+1}{3}-\frac{n}{3}=\frac{1}{3}$，
∴${a}_{n}=\frac{1}{{3}^{n}}$（n≥2）．
又由${a_1}+3{a_2}+{3^2}{a_3}+…+{3^{n-1}}{a_n}=\frac{n+1}{3}$，得${a}_{1}=\frac{2}{3}$不適合上式．
∴a_n=$\left\{\begin{array}{l}{\frac{2}{3}，n=1}\\{\frac{1}{{3}^{n}}，n≥2}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{\frac{2}{3}，n=1}\\{\frac{1}{{3}^{n}}，n≥2}\end{array}\right.$．

點(diǎn)評 本題考查數(shù)列遞推式，訓(xùn)練了作差法求數(shù)列的通項(xiàng)公式，是中檔題．