Question

4．已知數(shù)列{a_n}滿足a_n+1=$\left\{\begin{array}{l}2{a_n}（0≤{a_n}＜\frac{1}{2}）\\ 2{a_n}-1（\frac{1}{2}≤{a_n}＜1）\end{array}\right.$，若a₁=$\frac{6}{7}$，則a₂₀₁₇=（　�。�

• A． $\frac{1}{7}$
• B． $\frac{3}{7}$
• C． $\frac{5}{7}$
• D． $\frac{6}{7}$

Answer 1

分析 數(shù)列{a_n}滿足a_n+1=$\left\{\begin{array}{l}2{a_n}（0≤{a_n}＜\frac{1}{2}）\\ 2{a_n}-1（\frac{1}{2}≤{a_n}＜1）\end{array}\right.$，a₁=$\frac{6}{7}$，可得a_n+3=a_n．即可得出．

解答 解：∵數(shù)列{a_n}滿足a_n+1=$\left\{\begin{array}{l}2{a_n}（0≤{a_n}＜\frac{1}{2}）\\ 2{a_n}-1（\frac{1}{2}≤{a_n}＜1）\end{array}\right.$，a₁=$\frac{6}{7}$，
∴a₂=2a₁-1=$\frac{5}{7}$，a₃=2a₂-1=$\frac{3}{7}$，a₄=2a₃=$\frac{6}{7}$，…，
∴a_n+3=a_n．
則a₂₀₁₇=a_672×3+1=${a}_{1}=\frac{6}{7}$．
故選：D．

點(diǎn)評(píng) 本題考查了數(shù)列的遞推關(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．