新高考数列题真题及答案

新高考数列题真题及答案

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\(a_{5}\)的值为（）A.30B.31C.32D.33答案：B2.等比数列\(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，则公比\(q\)为（）A.\(-2\)B.2C.\(\pm2\)D.4答案：B3.已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(a_{3}=5\)，\(S_{3}=9\)，则\(a_{1}\)为（）A.1B.2C.3D.4答案：A4.数列\(\{a_{n}\}\)的通项公式\(a_{n}=n^{2}-5n+4\)，则数列\(\{a_{n}\}\)的最小项是（）A.第2项B.第3项C.第2项或第3项D.第4项答案：C5.设\(\{a_{n}\}\)是公差为\(d\)的等差数列，\(S_{n}\)为其前\(n\)项和，若\(S_{10}=S_{11}\)，且\(a_{1}>0\)，则（）A.\(d>0\)B.\(a_{11}=0\)C.\(S_{n}\)的最大值是\(S_{20}\)D.\(S_{n}\)的最小值是\(S_{20}\)答案：B6.已知等比数列\(\{a_{n}\}\)的各项均为正数，且\(a_{3}a_{5}=4\)，则\(\log_{2}a_{1}+\log_{2}a_{2}+\log_{2}a_{3}+\log_{2}a_{4}+\log_{2}a_{5}\)等于（）A.5B.10C.15D.20答案：A7.数列\(\{a_{n}\}\)满足\(a_{n+1}=\frac{1}{1-a_{n}}\)，\(a_{1}=2\)，则\(a_{2023}\)的值为（）A.2B.\(-1\)C.\(\frac{1}{2}\)D.1答案：A8.已知等差数列\(\{a_{n}\}\)中，\(a_{1}+a_{7}+a_{13}=3\pi\)，则\(\cos(a_{2}+a_{12})\)的值为（）A.\(\frac{1}{2}\)B.\(-\frac{1}{2}\)C.\(\frac{\sqrt{3}}{2}\)D.\(-\frac{\sqrt{3}}{2}\)答案：B9.等比数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，若\(S_{3}=1\)，\(S_{6}=9\)，则\(S_{9}\)等于（）A.27B.28C.29D.30答案：B10.已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=n^{2}-9n\)，第\(k\)项满足\(5<a_{k}<8\)，则\(k\)的值为（）A.6B.7C.8D.9答案：C二、多项选择题1.下列关于等差数列的说法中，正确的有（）A.若\(a,b,c\)成等差数列，则\(2b=a+c\)B.若\(a_{n}\)是等差数列，则\(a_{n}=kn+b\)（\(k,b\)为常数）C.等差数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=An^{2}+Bn\)（\(A,B\)为常数）D.若\(a_{n}\)是等差数列，\(m,n,p,q\inN^+\)，且\(m+n=p+q\)，则\(a_{m}+a_{n}=a_{p}+a_{q}\)答案：ABCD2.已知等比数列\(\{a_{n}\}\)的公比为\(q\)，下列说法正确的是（）A.若\(a_{1}>0\)，\(0<q<1\)，则数列\(\{a_{n}\}\)是递减数列B.若\(a_{1}<0\)，\(0<q<1\)，则数列\(\{a_{n}\}\)是递增数列C.若\(q>1\)，则数列\(\{a_{n}\}\)是递增数列D.若\(q<0\)，则数列\(\{a_{n}\}\)是摆动数列答案：ABD3.等差数列\(\{a_{n}\}\)的公差为\(d\)，前\(n\)项和为\(S_{n}\)，当首项\(a_{1}\)和\(d\)变化时，\(a_{3}+a_{8}+a_{13}\)是一个定值，则下列各数中为定值的有（）A.\(a_{7}\)B.\(a_{8}\)C.\(S_{15}\)D.\(S_{16}\)答案：BC4.已知数列\(\{a_{n}\}\)满足\(a_{n+1}=\frac{a_{n}}{1+a_{n}}\)，\(a_{1}=1\)，则下列说法正确的是（）A.数列\(\{a_{n}\}\)的通项公式\(a_{n}=\frac{1}{n}\)B.数列\(\{\frac{1}{a_{n}}\}\)是等差数列C.数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=\frac{n(n+1)}{2}\)D.数列\(\{a_{n}\}\)是递减数列答案：ABD5.等比数列\(\{a_{n}\}\)中，\(a_{3}=4\)，\(a_{7}=16\)，则（）A.\(a_{5}=8\)B.\(a_{5}=\pm8\)C.\(a_{1}=2\)D.\(a_{1}=\pm2\)答案：AC6.已知数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=n^{2}-n\)，则（）A.\(a_{1}=0\)B.数列\(\{a_{n}\}\)是等差数列C.\(a_{n}=2n-2\)D.\(S_{n}\)的最小值为\(-1\)答案：ABC7.设等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，公差为\(d\)，若\(S_{5}<S_{6}\)，\(S_{6}=S_{7}>S_{8}\)，则（）A.\(d<0\)B.\(a_{7}=0\)C.\(S_{9}>S_{5}\)D.\(S_{6}\)与\(S_{7}\)均为\(S_{n}\)的最大值答案：ABD8.已知数列\(\{a_{n}\}\)是等比数列，\(a_{3}\)，\(a_{7}\)是方程\(x^{2}-6x+4=0\)的两根，则\(a_{5}\)的值为（）A.\(\sqrt{6}\)B.\(-\sqrt{6}\)C.2D.\(-2\)答案：C9.对于数列\(\{a_{n}\}\)，定义\(H_{n}=\frac{a_{1}+2a_{2}+\cdots+2^{n-1}a_{n}}{n}\)为\(\{a_{n}\}\)的“优值”，已知某数列\(\{a_{n}\}\)的“优值”\(H_{n}=2^{n+1}\)，则（）A.\(a_{1}=4\)B.数列\(\{a_{n}\}\)是等差数列C.\(a_{n}=2n+2\)D.数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=n^{2}+3n\)答案：AC10.已知数列\(\{a_{n}\}\)满足\(a_{n+2}=a_{n+1}-a_{n}\)，\(a_{1}=1\)，\(a_{2}=2\)，则（）A.\(a_{2023}=-1\)B.数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)具有周期性C.\(a_{1}+a_{2}+\cdots+a_{2023}=1\)D.\(a_{6k+1}+a_{6k+2}+\cdots+a_{6k+6}=0\)（\(k\inN\)）答案：ABCD三、判断题1.常数列一定是等差数列。（）答案：对2.常数列一定是等比数列。（）答案：错3.若\(a_{n}=n^{2}\)，则数列\(\{a_{n}\}\)是等差数列。（）答案：错4.等比数列\(\{a_{n}\}\)中，若\(a_{1}<0\)，\(q>1\)，则数列\(\{a_{n}\}\)是递减数列。（）答案：对5.等差数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)，\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等差数列。（）答案：对6.等比数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)，\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等比数列。（）答案：错7.若数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}=3n^{2}-2n\)，则\(a_{n}=6n-5\)。（）答案：对8.若\(a,b,c\)成等比数列，则\(b^{2}=ac\)，反之也成立。（）答案：错9.等差数列\(\{a_{n}\}\)中，若\(a_{m}=n\)，\(a_{n}=m\)（\(m\neqn\)），则\(a_{m+n}=0\)。（）答案：对10.等比数列\(\{a_{n}\}\)中，\(a_{n}=a_{m}q^{n-m}\)（\(m,n\inN^+\)）。（）答案：对四、简答题1.已知等差数列\(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{7}=13\)，求数列\(\{a_{n}\}\)的通项公式\(a_{n}\)。答案：设等差数列\(\{a_{n}\}\)的公差为\(d\)。由\(a_{n}=a_{1}+(n-1)d\)，可得\(a_{3}=a_{1}+2d=5\)，\(a_{7}=a_{1}+6d=13\)。用\(a_{7}\)的式子减去\(a_{3}\)的式子，即\((a_{1}+6d)-(a_{1}+2d)=13-5\)，\(4d=8\)，解得\(d=2\)。把\(d=2\)代入\(a_{1}+2d=5\)，得\(a_{1}=1\)。所以\(a_{n}=1+(n-1)\times2=2n-1\)。2.已知等比数列\(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，求数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)。答案：设等比数列\(\{a_{n}\}\)的公比为\(q\)。由\(a_{n}=a_{1}q^{n-1}\)，可得\(a_{4}=a_{1}q^{3}\)，即\(16=2q^{3}\)，解得\(q=2\)。等比数列的前\(n\)项和公式为\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)）。把\(a_{1}=2\)，\(q=2\)代入可得\(S_{n}=\frac{2(1-2^{n})}{1-2}=2^{n+1}-2\)。3.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\(a_{n}\)。答案：当\(n=1\)时，\(a_{1}=S_{1}=1^{2}+1=2\)；当\(n\geq2\)时，\(a_{n}=S_{n}-S_{n-1}=(n^{2}+n)-[(n-1)^{2}+(n-1)]\)，展开得\(a_{n}=n^{2}+n-(n^{2}-2n+1+n-1)=2n\)。当\(n=1\)时，\(a_{1}=2\)也满足\(a_{n}=2n\)。所以数列\(\{a_{n}\}\)的通项公式\(a_{n}=2n\)。4.已知等差数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，