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山东名校考试联盟 2023-2024学年高二年级上学期期中检测(2023.11)数学考试卷试卷答案
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选项中有两项为多余答案选项
六、语法填空(共10小题:每小题1分,满分10分)Puttingafewofthesetipsintodailypracticecanhelpyoubecomeabetterlearner阅读下面短文,在空白处填人1个适当的单词或括号内单词的正确形式
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分析(1)运用数列极限公式$\underset{lim}{n→∞}$$\frac{1}{{2}^{n-1}}$=0;(2)运用等比数列的求和公式求得Sn,再取极限,即可得到所求值.
解答解:由an=$\left\{\begin{array}{l}{\frac{1}{n(n+1)},1≤n≤3}\\{\frac{1}{{2}^{n-1}}.n≥4}\end{array}\right.$,
(1)$\underset{lim}{n→∞}{a}_{n}$=$\underset{lim}{n→∞}$$\frac{1}{{2}^{n-1}}$=0;
(2)Sn为前n项的和,
即有Sn=$\frac{1}{2}$+$\frac{1}{6}$+$\frac{1}{12}$+$\frac{1}{8}$+$\frac{1}{16}$+…+$\frac{1}{{2}^{n-1}}$
=$\frac{3}{4}$+$\frac{\frac{1}{8}(1-\frac{1}{{2}^{n-3}})}{1-\frac{1}{2}}$=1-$\frac{1}{{2}^{n-1}}$,
即有$\underset{lim}{n→∞}{S}_{n}$=$\underset{lim}{n→∞}$(1-$\frac{1}{{2}^{n-1}}$)
=1-$\underset{lim}{n→∞}$$\frac{1}{{2}^{n-1}}$=1-0=1.
点评本题考查等比数列的通项和求和公式的运用,考查数列极限的求法,考查运算能力,属于中档题.