Question

3．已知數(shù)列{a_n}中${a_1}=2，{a_2}=1，{a_{n+2}}=\left\{\begin{array}{l}\frac{{2{a_{n+1}}}}{a_n}，{a_{n+1}}≥2\\ \frac{4}{a_n}，{a_{n+1}}＜2\end{array}\right.（n∈{N^*}），{S_n}$是數(shù)列{a_n}的前n項(xiàng)和，則S₂₀₁₆=5241．

Answer 1

分析 由a₁=2，a₂=1，a_n+2=$\left\{\begin{array}{l}{\frac{2{a}_{n+1}}{{a}_{n}}，{a}_{n+1}≥2}\\{\frac{4}{{a}_{n}}，{a}_{n+1}＜2}\end{array}\right.$．可得a_n+5=a_n．即可得出．

解答 解：∵a₁=2，a₂=1，a_n+2=$\left\{\begin{array}{l}{\frac{2{a}_{n+1}}{{a}_{n}}，{a}_{n+1}≥2}\\{\frac{4}{{a}_{n}}，{a}_{n+1}＜2}\end{array}\right.$．
∴a₃=$\frac{4}{{a}_{1}}$=2，a₄=$\frac{2{a}_{3}}{{a}_{2}}$=4，a₅=4，a₆=2，a₇=1，…，
∴a_n+5=a_n．
∴S₂₀₁₆=S_5×403+1=（2+1+2+4+4）×403+2=5241．
故答案為：5241．

點(diǎn)評 本題考查了遞推關(guān)系的應(yīng)用、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．