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4. 嫦娥二号卫星在完成探月任务后, 继续进行深空探测, 成为我国第一颗环绕太阳飞行的 人造行星, 为研究嫦娥二号绕日周期与地球绕日周期的比值, 用到数列 $\left\{b_{n}\right\}: b_{1}=1+\frac{1}{a_{1}}$, $b_{2}=1+\frac{1}{a_{1}+\frac{1}{a_{2}}}, \quad b_{3}=1+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}}}}, \cdots$, 依此类推, 其中 $a_{k} \in \mathbf{N}^{\star}(k=1,2, \cdots)$. 则 ()

Gaokao · Math · previous-year question

  1. A.$b_{1}<b_{5}$
  2. B.$b_{3}<b_{8}$
  3. C.$b_{6}<b_{2}$
  4. D.$b_{4}<b_{7}$correct

Answer

D. $b_{4}<b_{7}$

Explanation

【详解】解: 因为 $a_{k} \in \mathbf{N}^{*}(k=1,2, \cdots)$, 所以 $a_{1}<a_{1}+\frac{1}{a_{2}}, \frac{1}{a_{1}}>\frac{1}{a_{1}+\frac{1}{a_{2}}}$, 得到 $b_{1}>b_{2}$, 同理 $a_{1}+\frac{1}{a_{2}}>a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}}}$, 可得 $b_{2}<b_{3}, b_{1}>b_{3}$ 又因为 $\frac{1}{a_{2}}>\frac{1}{a_{2}+\frac{1}{a_{3}+\frac{1}{a_{4}}}}, a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}}}<a_{1}+\frac{1}{a_{2}+\frac{1}{a_{3}+\frac{1}{a_{4}}}}$ , 故 $b_{2}<b_{4}, b_{3}>b_{4}$; 以此类推, 可得 $b_{1}>b_{3}>b_{5}>b_{7}>\cdots, b_{7}>b_{8}$, 故 $\mathrm{A}$ 错误; $b_{1}>b_{7}>b_{8}$, 故 B 错误; $$ & \frac{1}{a_{2}}>\frac{1}{a_{2}

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