高考数列放缩题目及答案

高考数列放缩题目及答案

一、单项选择题1.已知数列\(\{a_{n}\}\)满足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，则\(a_{n}\)的通项公式为（）A.\(a_{n}=2^{n}-1\)B.\(a_{n}=2^{n-1}\)C.\(a_{n}=2n-1\)D.\(a_{n}=2^{n}+1\)答案：A2.数列\(\{a_{n}\}\)中，\(a_{n}=\frac{1}{n(n+1)}\)，其前\(n\)项和\(S_{n}\)为（）A.\(\frac{n}{n+1}\)B.\(\frac{1}{n+1}\)C.\(\frac{n+1}{n}\)D.\(\frac{n}{n-1}\)答案：A3.设数列\(\{a_{n}\}\)是等差数列，若\(a_{3}+a_{4}+a_{5}=12\)，则\(a_{1}+a_{2}+\cdots+a_{7}\)等于（）A.14B.21C.28D.35答案：C4.已知等比数列\(\{a_{n}\}\)的公比\(q=2\)，前\(n\)项和为\(S_{n}\)，若\(S_{4}=1\)，则\(S_{8}\)等于（）A.15B.17C.19D.21答案：B5.数列\(\{a_{n}\}\)满足\(a_{n}=2^{n-1}\)，\(b_{n}=\log_{2}a_{n}\)，则数列\(\{b_{n}\}\)的前\(n\)项和\(T_{n}\)为（）A.\(\frac{n(n-1)}{2}\)B.\(\frac{n(n+1)}{2}\)C.\(n(n-1)\)D.\(n(n+1)\)答案：A6.若数列\(\{a_{n}\}\)的通项公式\(a_{n}=(-1)^{n}(3n-2)\)，则\(a_{1}+a_{2}+\cdots+a_{10}\)等于（）A.15B.12C.-12D.-15答案：A7.已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=n^{2}-9n\)，第\(k\)项满足\(5\lta_{k}\lt8\)，则\(k\)等于（）A.9B.8C.7D.6答案：B8.数列\(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{n+1}-a_{n}=2^{n}\)，则\(a_{n}\)等于（）A.\(2^{n}-1\)B.\(2^{n-1}\)C.\(2^{n+1}-3\)D.\(2^{n+1}-1\)答案：A9.设等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{3}=9\)，\(S_{6}=36\)，则\(a_{7}+a_{8}+a_{9}\)等于（）A.63B.45C.36D.27答案：B10.已知数列\(\{a_{n}\}\)满足\(a_{1}=1\)，\(a_{n+1}=\frac{a_{n}}{1+a_{n}}\)，则\(a_{n}\)等于（）A.\(\frac{1}{n}\)B.\(\frac{1}{n+1}\)C.\(\frac{1}{2n-1}\)D.\(\frac{1}{2^{n-1}}\)答案：A二、多项选择题1.设\(\{a_{n}\}\)是等差数列，\(S_{n}\)是其前\(n\)项和，且\(S_{5}\ltS_{6}\)，\(S_{6}=S_{7}\gtS_{8}\)，则下列结论正确的是（）A.\(d\lt0\)B.\(a_{7}=0\)C.\(S_{9}\gtS_{5}\)D.\(S_{6}\)与\(S_{7}\)均为\(S_{n}\)的最大值答案：ABD2.已知等比数列\(\{a_{n}\}\)中，\(a_{5}=3\)，\(a_{15}=27\)，则\(a_{10}\)可能为（）A.\(-9\)B.9C.\(\sqrt{81}\)D.\(-\sqrt{81}\)答案：AB3.数列\(\{a_{n}\}\)满足\(a_{n+1}=\frac{1}{1-a_{n}}\)，\(a_{8}=2\)，则（）A.\(a_{1}=\frac{1}{2}\)B.\(a_{2}=-1\)C.\(a_{3}=2\)D.\(a_{n}\)是周期数列答案：ABCD4.对于数列\(\{a_{n}\}\)，若存在正整数\(T\)，对于任意正整数\(n\)都有\(a_{n+T}=a_{n}\)成立，则称数列\(\{a_{n}\}\)是以\(T\)为周期的周期数列。设\(b_{1}=m(0\ltm\lt1)\)，对任意正整数\(n\)都有\(b_{n+1}=\begin{cases}2b_{n},&0\leqb_{n}\lt\frac{1}{2}\\2-2b_{n},&\frac{1}{2}\leqb_{n}\lt1\end{cases}\)，若数列\(\{b_{n}\}\)是以\(3\)为周期的周期数列，则\(m\)的值可以为（）A.\(\frac{1}{4}\)B.\(\frac{1}{3}\)C.\(\frac{5}{8}\)D.\(\frac{3}{4}\)答案：BC5.已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=2a_{n}-1\)，则下列结论正确的是（）A.\(a_{1}=1\)B.\(a_{n}=2^{n-1}\)C.\(a_{n}=2n-1\)D.\(S_{n}=2^{n}-1\)答案：ABD6.设等差数列\(\{a_{n}\}\)的公差为\(d\)，前\(n\)项和为\(S_{n}\)，若\(a_{3}=12\)，\(S_{12}\gt0\)，\(S_{13}\lt0\)，则（）A.\(d\lt0\)B.\(a_{7}\gt0\)C.\(S_{5}=60\)D.\(a_{6}\gt0\)答案：ACD7.已知数列\(\{a_{n}\}\)满足\(a_{n+1}-a_{n}=2n\)，\(a_{1}=1\)，则（）A.\(a_{n}=n^{2}-n+1\)B.\(a_{n}=n^{2}+n-1\)C.数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=\frac{n(n^{2}+1)}{3}\)D.数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=\frac{n(n^{2}-1)}{3}\)答案：AC8.等比数列\(\{a_{n}\}\)的公比为\(q\)，其前\(n\)项积为\(T_{n}\)，并且满足条件\(a_{1}\gt1\)，\(a_{99}a_{100}-1\gt0\)，\(\frac{a_{99}-1}{a_{100}-1}\lt0\)，则下列结论正确的是（）A.\(0\ltq\lt1\)B.\(a_{99}a_{101}-1\lt0\)C.\(T_{100}\)是\(T_{n}\)中的最大值D.使\(T_{n}\gt1\)成立的最大自然数\(n\)等于\(198\)答案：ABD9.已知数列\(\{a_{n}\}\)的通项公式\(a_{n}=\frac{n}{n^{2}+90}\)，则（）A.\(a_{n}\)的最大值为\(\frac{1}{19}\)B.\(a_{n}\)的最大值为\(\frac{1}{18}\)C.\(n=9\)或\(n=10\)时\(a_{n}\)取得最大值D.\(n=1\)时\(a_{n}\)取得最大值答案：AC10.设数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{n}=n^{2}-2n\)，则（）A.\(a_{n}=2n-3\)B.\(a_{n}=2n+3\)C.数列\(\{a_{n}\}\)是等差数列D.数列\(\{a_{n}\}\)是等比数列答案：AC三、判断题1.若数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=n^{2}+1\)，则\(a_{n}=2n-1\)。（）答案：错误2.等差数列\(\{a_{n}\}\)中，若\(a_{1}+a_{5}=10\)，\(a_{4}=7\)，则公差\(d=2\)。（）答案：正确3.等比数列\(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{4}=8\)，则公比\(q=2\)。（）答案：正确4.数列\(\{a_{n}\}\)满足\(a_{n+1}=2a_{n}\)，\(a_{1}=1\)，则\(a_{n}=2^{n}\)。（）答案：错误5.若数列\(\{a_{n}\}\)的通项公式\(a_{n}=n^{2}-n\)，则\(\{a_{n}\}\)是递增数列。（）答案：正确6.等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{3}=6\)，\(a_{1}=4\)，则公差\(d=-2\)。（）答案：正确7.等比数列\(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，则公比\(q=2\)。（）答案：正确8.数列\(\{a_{n}\}\)满足\(a_{n+1}-a_{n}=n\)，\(a_{1}=1\)，则\(a_{n}=\frac{n(n-1)}{2}+1\)。（）答案：正确9.若数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=3^{n}-1\)，则\(a_{n}=2\cdot3^{n-1}\)。（）答案：正确10.等差数列\(\{a_{n}\}\)中，若\(a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=450\)，则\(a_{2}+a_{8}=180\)。（）答案：正确四、简答题1.已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=n^{2}-n\)，求数列\(\{a_{n}\}\)的通项公式。答案：当\(n=1\)时，\(a_{1}=S_{1}=1^{2}-1=0\)；当\(n\geq2\)时，\(a_{n}=S_{n}-S_{n-1}=n^{2}-n-[(n-1)^{2}-(n-1)]\)，化简得\(a_{n}=2n-2\)。当\(n=1\)时，\(a_{1}=0\)也满足\(a_{n}=2n-2\)。所以数列\(\{a_{n}\}\)的通项公式为\(a_{n}=2n-2\)。2.已知等差数列\(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{5}=9\)，求数列\(\{a_{n}\}\)的通项公式及前\(n\)项和\(S_{n}\)。答案：设等差数列\(\{a_{n}\}\)的公差为\(d\)，则\(2d=a_{5}-a_{3}=9-5=4\)，解得\(d=2\)。由\(a_{3}=a_{1}+2d=5\)，可得\(a_{1}=5-2\times2=1\)。所以\(a_{n}=a_{1}+(n-1)d=1+2(n-1)=2n-1\)。\(S_{n}=na_{1}+\frac{n(n-1)}{2}d=n\times1+\frac{n(n-1)}{2}\times2=n^{2}\)。3.已知等比数列\(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，求数列\(\{a_{n}\}\)的通项公式及前\(n\)项和\(S_{n}\)。答案：设等比数列\(\{a_{n}\}\)的公比为\(q\)，由\(a_{4}=a_{1}q^{3}\)，即\(16=2q^{3}\)，解得\(q=2\)。所