新高考數(shù)列題真題及答案

一、單項選擇題1.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，則\(a_{5}\)的值為（）A.30B.31C.32D.33答案：B2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q\)為（）A.\(-2\)B.2C.\(\pm2\)D.4答案：B3.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(a_{3}=5\)，\(S_{3}=9\)，則\(a_{1}\)為（）A.1B.2C.3D.4答案：A4.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=n^{2}-5n+4\)，則數(shù)列\(zhòng)(\{a_{n}\}\)的最小項是（）A.第2項B.第3項C.第2項或第3項D.第4項答案：C5.設\(\{a_{n}\}\)是公差為\(d\)的等差數(shù)列，\(S_{n}\)為其前\(n\)項和，若\(S_{10}=S_{11}\)，且\(a_{1}>0\)，則（）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.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的各項均為正數(shù)，且\(a_{3}a_{5}=4\)，則\(\log_{2}a_{1}+\log_{2}a_{2}+\log_{2}a_{3}+\log_{2}a_{4}+\log_{2}a_{5}\)等于（）A.5B.10C.15D.20答案：A7.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\frac{1}{1-a_{n}}\)，\(a_{1}=2\)，則\(a_{2023}\)的值為（）A.2B.\(-1\)C.\(\frac{1}{2}\)D.1答案：A8.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{7}+a_{13}=3\pi\)，則\(\cos(a_{2}+a_{12})\)的值為（）A.\(\frac{1}{2}\)B.\(-\frac{1}{2}\)C.\(\frac{\sqrt{3}}{2}\)D.\(-\frac{\sqrt{3}}{2}\)答案：B9.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{3}=1\)，\(S_{6}=9\)，則\(S_{9}\)等于（）A.27B.28C.29D.30答案：B10.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-9n\)，第\(k\)項滿足\(5<a_{k}<8\)，則\(k\)的值為（）A.6B.7C.8D.9答案：C二、多項選擇題1.下列關于等差數(shù)列的說法中，正確的有（）A.若\(a,b,c\)成等差數(shù)列，則\(2b=a+c\)B.若\(a_{n}\)是等差數(shù)列，則\(a_{n}=kn+b\)（\(k,b\)為常數(shù)）C.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=An^{2}+Bn\)（\(A,B\)為常數(shù)）D.若\(a_{n}\)是等差數(shù)列，\(m,n,p,q\inN^+\)，且\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)答案：ABCD2.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，下列說法正確的是（）A.若\(a_{1}>0\)，\(0<q<1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞減數(shù)列B.若\(a_{1}<0\)，\(0<q<1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列C.若\(q>1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列D.若\(q<0\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是擺動數(shù)列答案：ABD3.等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，前\(n\)項和為\(S_{n}\)，當首項\(a_{1}\)和\(d\)變化時，\(a_{3}+a_{8}+a_{13}\)是一個定值，則下列各數(shù)中為定值的有（）A.\(a_{7}\)B.\(a_{8}\)C.\(S_{15}\)D.\(S_{16}\)答案：BC4.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\frac{a_{n}}{1+a_{n}}\)，\(a_{1}=1\)，則下列說法正確的是（）A.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=\frac{1}{n}\)B.數(shù)列\(zhòng)(\{\frac{1}{a_{n}}\}\)是等差數(shù)列C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=\frac{n(n+1)}{2}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)是遞減數(shù)列答案：ABD5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=4\)，\(a_{7}=16\)，則（）A.\(a_{5}=8\)B.\(a_{5}=\pm8\)C.\(a_{1}=2\)D.\(a_{1}=\pm2\)答案：AC6.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-n\)，則（）A.\(a_{1}=0\)B.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列C.\(a_{n}=2n-2\)D.\(S_{n}\)的最小值為\(-1\)答案：ABC7.設等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，公差為\(d\)，若\(S_{5}<S_{6}\)，\(S_{6}=S_{7}>S_{8}\)，則（）A.\(d<0\)B.\(a_{7}=0\)C.\(S_{9}>S_{5}\)D.\(S_{6}\)與\(S_{7}\)均為\(S_{n}\)的最大值答案：ABD8.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，\(a_{3}\)，\(a_{7}\)是方程\(x^{2}-6x+4=0\)的兩根，則\(a_{5}\)的值為（）A.\(\sqrt{6}\)B.\(-\sqrt{6}\)C.2D.\(-2\)答案：C9.對于數(shù)列\(zhòng)(\{a_{n}\}\)，定義\(H_{n}=\frac{a_{1}+2a_{2}+\cdots+2^{n-1}a_{n}}{n}\)為\(\{a_{n}\}\)的“優(yōu)值”，已知某數(shù)列\(zhòng)(\{a_{n}\}\)的“優(yōu)值”\(H_{n}=2^{n+1}\)，則（）A.\(a_{1}=4\)B.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列C.\(a_{n}=2n+2\)D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}+3n\)答案：AC10.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+2}=a_{n+1}-a_{n}\)，\(a_{1}=1\)，\(a_{2}=2\)，則（）A.\(a_{2023}=-1\)B.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)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、判斷題1.常數(shù)列一定是等差數(shù)列。（）答案：對2.常數(shù)列一定是等比數(shù)列。（）答案：錯3.若\(a_{n}=n^{2}\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列。（）答案：錯4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}<0\)，\(q>1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞減數(shù)列。（）答案：對5.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)，\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等差數(shù)列。（）答案：對6.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)，\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等比數(shù)列。（）答案：錯7.若數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=3n^{2}-2n\)，則\(a_{n}=6n-5\)。（）答案：對8.若\(a,b,c\)成等比數(shù)列，則\(b^{2}=ac\)，反之也成立。（）答案：錯9.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{m}=n\)，\(a_{n}=m\)（\(m\neqn\)），則\(a_{m+n}=0\)。（）答案：對10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}=a_{m}q^{n-m}\)（\(m,n\inN^+\)）。（）答案：對四、簡答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{7}=13\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}\)。答案：設等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)。由\(a_{n}=a_{1}+(n-1)d\)，可得\(a_{3}=a_{1}+2d=5\)，\(a_{7}=a_{1}+6d=13\)。用\(a_{7}\)的式子減去\(a_{3}\)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.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)。答案：設等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)。由\(a_{n}=a_{1}q^{n-1}\)，可得\(a_{4}=a_{1}q^{3}\)，即\(16=2q^{3}\)，解得\(q=2\)。等比數(shù)列的前\(n\)項和公式為\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)）。把\(a_{1}=2\)，\(q=2\)代入可得\(S_{n}=\frac{2(1-2^{n})}{1-2}=2^{n+1}-2\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}+n\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}\)。答案：當\(n=1\)時，\(a_{1}=S_{1}=1^{2}+1=2\)；當\(n\geq2\)時，\(a_{n}=S_{n}-S_{n-1}=(n^{2}+n)-[(n-1)^{2}+(n-1)]\)，展開得\(a_{n}=n^{2}+n-(n^{2}-2n+1+n-1)=2n\)。當\(n=1\)時，\(a_{1}=2\)也滿足\(a_{n}=2n\)。所以數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=2n\)。4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，