新高考試卷數(shù)列題及答案

一、單項選擇題（每題2分，共10題）1.數(shù)列\(zhòng)(1,3,5,7,\cdots\)的通項公式是（）A.\(a_{n}=2n-1\)B.\(a_{n}=2n+1\)C.\(a_{n}=n+1\)D.\(a_{n}=n-1\)2.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{3}=6\)，則公差\(d\)為（）A.1B.2C.3D.43.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{2}=2\)，則公比\(q\)是（）A.1B.2C.3D.44.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(a_{5}\)的值為（）A.9B.11C.15D.175.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(a_{3}=5\)，則\(S_{5}\)等于（）A.15B.20C.25D.306.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}a_{5}=16\)，則\(a_{4}\)的值為（）A.4B.\(\pm4\)C.8D.\(\pm8\)7.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2\)，\(a_{1}=1\)，則\(a_{10}\)為（）A.17B.19C.21D.238.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)，\(a_{2}+a_{8}=10\)，則\(a_{5}\)的值是（）A.5B.6C.8D.109.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，則\(q^{3}\)的值為（）A.4B.8C.16D.3210.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=\frac{1}{n(n+1)}\)，則\(a_{3}\)的值是（）A.\(\frac{1}{6}\)B.\(\frac{1}{12}\)C.\(\frac{1}{20}\)D.\(\frac{1}{30}\)二、多項選擇題（每題2分，共10題）1.以下哪些是等差數(shù)列（）A.\(1,2,3,4,\cdots\)B.\(2,4,8,16,\cdots\)C.\(5,5,5,5,\cdots\)D.\(1,3,9,27,\cdots\)2.等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q\)可以是（）A.1B.-1C.0D.23.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}=1\)，\(d=2\)，則（）A.\(a_{2}=3\)B.\(a_{3}=5\)C.\(a_{4}=7\)D.\(a_{5}=9\)4.對于數(shù)列\(zhòng)(\{a_{n}\}\)，\(S_{n}\)是其前\(n\)項和，下列說法正確的是（）A.\(a_{1}=S_{1}\)B.\(a_{n}=S_{n}-S_{n-1}(n\geq2)\)C.\(S_{n}=a_{1}+a_{2}+\cdots+a_{n}\)D.若\(S_{n}=n^{2}\)，則\(a_{n}=2n-1\)5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=3\)，則（）A.\(a_{2}=3\)B.\(a_{3}=9\)C.\(a_{4}=27\)D.\(a_{5}=81\)6.下列數(shù)列通項公式中，哪些能表示有窮數(shù)列（）A.\(a_{n}=n(n\leq5)\)B.\(a_{n}=2^{n}\)C.\(a_{n}=\frac{1}{n}(n=1,2,3)\)D.\(a_{n}=n^{2}+1\)7.等差數(shù)列\(zhòng)(\{a_{n}\}\)的性質(zhì)有（）A.若\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)B.\(a_{n}=a_{m}+(n-m)d\)C.\(S_{n}=\frac{n(a_{1}+a_{n})}{2}\)D.連續(xù)\(m\)項的和\(S_{m},S_{2m}-S_{m},S_{3m}-S_{2m}\)仍成等差數(shù)列8.等比數(shù)列\(zhòng)(\{a_{n}\}\)的性質(zhì)有（）A.若\(m+n=p+q\)，則\(a_{m}a_{n}=a_{p}a_{q}\)B.\(a_{n}=a_{m}q^{n-m}\)C.\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}(q\neq1)\)D.連續(xù)\(m\)項的積\(T_{m},\frac{T_{2m}}{T_{m}},\frac{T_{3m}}{T_{2m}}\)仍成等比數(shù)列9.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=3\)，\(a_{1}=1\)，則（）A.\(\{a_{n}\}\)是等差數(shù)列B.\(a_{n}=3n-2\)C.\(S_{n}=\frac{n(1+3n-2)}{2}\)D.\(a_{5}=13\)10.數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，\(a_{2}=2\)，\(a_{4}=8\)，則（）A.\(q^{2}=4\)B.\(q=2\)C.\(a_{1}=1\)D.\(a_{3}=\pm4\)三、判斷題（每題2分，共10題）1.常數(shù)列一定是等差數(shù)列。（）2.常數(shù)列一定是等比數(shù)列。（）3.若\(a_{n}=3n\)，則\(\{a_{n}\}\)是等差數(shù)列。（）4.若\(a_{n}=2^{n}\)，則\(\{a_{n}\}\)是等比數(shù)列。（）5.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)是前\(n\)項和，則\(S_{n}\)，\(S_{2n}\)，\(S_{3n}\)成等差數(shù)列。（）6.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)是前\(n\)項和，則\(S_{n}\)，\(S_{2n}\)，\(S_{3n}\)成等比數(shù)列。（）7.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=n^{2}+1\)，則\(a_{3}=10\)。（）8.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=2a_{n}\)，\(a_{1}=1\)，則\(a_{n}=2^{n-1}\)。（）9.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}=0\)，\(d=1\)，則\(a_{n}=n-1\)。（）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=-1\)，則\(a_{n}=(-1)^{n-1}\)。（）四、簡答題（每題5分，共4題）1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=3\)，\(d=2\)，求\(a_{n}\)。答案：根據(jù)等差數(shù)列通項公式\(a_{n}=a_{1}+(n-1)d\)，將\(a_{1}=3\)，\(d=2\)代入，得\(a_{n}=3+2(n-1)=2n+1\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(q=3\)，求\(S_{4}\)。答案：由等比數(shù)列前\(n\)項和公式\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}(q\neq1)\)，把\(a_{1}=2\)，\(q=3\)，\(n=4\)代入，\(S_{4}=\frac{2(1-3^{4})}{1-3}=80\)。3.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}+n\)，求\(a_{n}\)。答案：當\(n=1\)時，\(a_{1}=S_{1}=1^{2}+1=2\)；當\(n\geq2\)時，\(a_{n}=S_{n}-S_{n-1}=(n^{2}+n)-[(n-1)^{2}+(n-1)]=2n\)。\(n=1\)時也滿足，所以\(a_{n}=2n\)。4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=7\)，\(a_{5}=11\)，求\(a_{1}\)和\(d\)。答案：因為\(a_{n}=a_{1}+(n-1)d\)，則\(a_{3}=a_{1}+2d=7\)，\(a_{5}=a_{1}+4d=11\)，兩式相減得\(2d=4\)，\(d=2\)，把\(d=2\)代入\(a_{1}+2d=7\)，得\(a_{1}=3\)。五、討論題（每題5分，共4題）1.討論等差數(shù)列和等比數(shù)列在實際生活中的應(yīng)用。答案：等差數(shù)列在計算銀行存款利息（按單利計算）、樓層高度遞增等場景有應(yīng)用；等比數(shù)列常用于計算銀行復(fù)利、細胞分裂、病毒傳播等問題，它們能幫助我們分析和預(yù)測實際變化規(guī)律。2.探討如何根據(jù)數(shù)列的前幾項判斷數(shù)列類型。答案：若相鄰兩項差值恒定，可能是等差數(shù)列；若相鄰兩項比值恒定，可能是等比數(shù)列。還需多計算幾項驗證，同時觀察數(shù)列各項特征，結(jié)合通項公式形式輔助判斷，不能僅依據(jù)少數(shù)項確定。3.說說數(shù)列的通項公式和前\(n\)項和公式的聯(lián)系與區(qū)別。答案：聯(lián)系：可由通項公式求前\(n\)項和，也能根據(jù)前\(n\)項和求通項。區(qū)別：通項公式描述數(shù)列每項規(guī)律，前\(n\)項和公式是求前\(n\)項的總和，形式和用途不同，通項側(cè)重項，前\(n\)項和側(cè)重整體和。4.討論在數(shù)列中求最大項或最小項的方法。答案：對于等差數(shù)列，若公差\(d\gt0\)，首項最?。籠(d\lt0\)，首項最大。等比數(shù)列需結(jié)合公比和首項判斷。也可通過作差\(a_{n+1}-a_{n}\)或作商\(\frac{a_{n+1}}{a_{n}}\)判斷單調(diào)性，進而確定最大或最小