Question

7．?dāng)?shù)列{a_n}滿足${a_{n+1}}=\left\{{\begin{array}{l}{2{a_n}}\\{{a_n}-1}\end{array}}\right.\begin{array}{l}{（0≤{a_n}≤1）}\\{（{a_n}＞1）}\end{array}$，且${a_1}=\frac{6}{7}$，則a₂₀₁₇=$\frac{12}{7}$．

Answer 1

分析 ${a_{n+1}}=\left\{{\begin{array}{l}{2{a_n}}\\{{a_n}-1}\end{array}}\right.\begin{array}{l}{（0≤{a_n}≤1）}\\{（{a_n}＞1）}\end{array}$，且${a_1}=\frac{6}{7}$，可得a_n+5=a_n．利用周期性即可得出．

解答 解：∵${a_{n+1}}=\left\{{\begin{array}{l}{2{a_n}}\\{{a_n}-1}\end{array}}\right.\begin{array}{l}{（0≤{a_n}≤1）}\\{（{a_n}＞1）}\end{array}$，且${a_1}=\frac{6}{7}$，
∴a₂=2a₁=$\frac{12}{7}$，a₃=a₂-1=$\frac{5}{7}$，a₄=2a₃=$\frac{10}{7}$，a₅=a₄-1=$\frac{3}{7}$，a₆=2a₅=$\frac{6}{7}$，…，
∴a_n+5=a_n．
則a₂₀₁₇=a_403×5+2=a₂=$\frac{12}{7}$．
故答案為：$\frac{12}{7}$．

點(diǎn)評 本題考查了數(shù)列遞推關(guān)系、數(shù)列的周期性，考查了推理能力與計(jì)算能力，屬于中檔題．