求解數(shù)列高三題目及答案

單項選擇題（每題2分，共10題）1.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{3}=6\)，則公差\(d\)為（）A.1B.2C.3D.42.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{2}=2\)，則公比\(q\)是（）A.1B.2C.3D.43.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=3n-1\)，則\(a_{5}\)的值為（）A.14B.15C.16D.174.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(a_{2}\)等于（）A.1B.3C.5D.75.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=4\)，\(a_{5}=16\)，則\(a_{4}\)為（）A.8B.-8C.±8D.106.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2\)，\(a_{1}=1\)，則\(a_{4}\)為（）A.5B.7C.9D.117.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{3}=8\)，\(a_{2}=4\)，則\(a_{5}\)為（）A.8B.10C.12D.148.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{4}=8\)，則\(a_{2}\)是（）A.2B.-2C.±2D.49.數(shù)列\(zhòng)(\{a_{n}\}\)的通項\(a_{n}=\frac{1}{n(n+1)}\)，則\(a_{3}\)的值為（）A.\(\frac{1}{6}\)B.\(\frac{1}{12}\)C.\(\frac{1}{20}\)D.\(\frac{1}{30}\)10.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}=2n+1\)，則\(a_{1}\)與公差\(d\)分別是（）A.3，2B.2，3C.3，3D.2，2多項選擇題（每題2分，共10題）1.下列數(shù)列中，是等差數(shù)列的有（）A.\(1,3,5,7,\cdots\)B.\(2,4,8,16,\cdots\)C.\(5,5,5,5,\cdots\)D.\(1,\frac{3}{2},2,\frac{5}{2},\cdots\)2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，公比\(q\gt1\)，則（）A.\(a_{2}\gta_{1}\)B.\(a_{3}\gta_{2}\)C.數(shù)列遞增D.\(a_{1}a_{3}\gta_{2}^{2}\)3.對于數(shù)列\(zhòng)(\{a_{n}\}\)，其前\(n\)項和\(S_{n}=n^{2}-n\)，則（）A.\(a_{1}=0\)B.\(a_{2}=2\)C.數(shù)列是等差數(shù)列D.\(a_{n}=2n-2\)4.以下關(guān)于等差數(shù)列性質(zhì)正確的有（）A.若\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)B.\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)成等差數(shù)列C.公差\(d\gt0\)時，數(shù)列遞增D.奇數(shù)項與偶數(shù)項分別成等差數(shù)列5.等比數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{3}=4\)，則（）A.\(q=2\)B.\(q=-2\)C.\(a_{2}=2\)D.\(a_{2}=-2\)6.數(shù)列\(zhòng)(\{a_{n}\}\)通項\(a_{n}=2^{n}\)，則（）A.是等比數(shù)列B.\(a_{3}=8\)C.前\(n\)項和\(S_{n}=2^{n+1}-2\)D.\(a_{n+1}-a_{n}=2^{n}\)7.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=3n^{2}+2n\)，則（）A.\(a_{1}=5\)B.公差\(d=6\)C.\(a_{n}=6n-1\)D.\(S_{2}=16\)8.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}\gt0\)，\(a_{2}a_{4}=16\)，則（）A.\(a_{3}=4\)B.\(a_{1}a_{5}=16\)C.\(a_{3}^{2}=a_{2}a_{4}\)D.\(a_{4}=8\)9.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=3\)，\(a_{1}=1\)，則（）A.是等差數(shù)列B.\(a_{2}=4\)C.\(a_{n}=3n-2\)D.前\(n\)項和\(S_{n}=\frac{n(3n-1)}{2}\)10.以下數(shù)列中，通項公式可以為\(a_{n}=n^{2}\)的有（）A.\(1,4,9,16,\cdots\)B.\(0,1,4,9,\cdots\)C.\(1,9,25,49,\cdots\)D.\(4,9,16,25,\cdots\)判斷題（每題2分，共10題）1.常數(shù)列一定是等差數(shù)列。（）2.等比數(shù)列的公比可以為\(0\)。（）3.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(a_{n}=2n-1\)。（）4.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\lt0\)，\(d\gt0\)，則數(shù)列遞增。（）5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=-1\)，則\(a_{n}=(-1)^{n-1}\)。（）6.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=2a_{n}\)，則是等比數(shù)列。（）7.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)一定是關(guān)于\(n\)的二次函數(shù)。（）8.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}a_{6}=a_{4}^{2}\)。（）9.若數(shù)列\(zhòng)(\{a_{n}\}\)的通項\(a_{n}=n+\frac{1}{n}\)，則\(a_{n}\)是等差數(shù)列。（）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{3}=4\)，則\(a_{2}=2\)。（）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\(a_{n}\)與\(S_{n}\)。-答案：\(a_{n}=a_{1}+(n-1)d=3+2(n-1)=2n+1\)；\(S_{n}=na_{1}+\frac{n(n-1)}{2}d=3n+n(n-1)=n^{2}+2n\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(q=3\)，求\(a_{4}\)與\(S_{4}\)。-答案：\(a_{4}=a_{1}q^{3}=2×3^{3}=54\)；\(S_{4}=\frac{a_{1}(1-q^{4})}{1-q}=\frac{2(1-3^{4})}{1-3}=80\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}+n\)，求\(a_{n}\)。-答案：當(dāng)\(n=1\)時，\(a_{1}=S_{1}=1^{2}+1=2\)；當(dāng)\(n\geq2\)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.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{5}=9\)，求\(a_{1}\)與\(d\)。-答案：由\(a_{n}=a_{1}+(n-1)d\)可得\(\begin{cases}a_{1}+2d=5\\a_{1}+4d=9\end{cases}\)，兩式相減得\(2d=4\)，即\(d=2\)，將\(d=2\)代入\(a_{1}+2d=5\)得\(a_{1}=1\)。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)用實例及意義。-答案：等差數(shù)列如銀行零存整取，每月固定存相同金額，最終本息和計算用到等差數(shù)列知識。等比數(shù)列如細(xì)胞分裂、復(fù)利計算等。意義在于幫助解決經(jīng)濟(jì)、科學(xué)等領(lǐng)域的增長、變化規(guī)律問題，為決策提供依據(jù)。2.若數(shù)列\(zhòng)(\{a_{n}\}\)既是等差數(shù)列又是等比數(shù)列，這樣的數(shù)列有什么特點？-答案：非零常數(shù)列既是等差數(shù)列（公差為\(0\)）又是等比數(shù)列（公比為\(1\)）。因為若\(a_{n}\)是等差數(shù)列，\(a_{n+1}-a_{n}=d\)，是等比數(shù)列\(zhòng)(\frac{a_{n+1}}{a_{n}}=q\)，只有\(zhòng)(d=0\)且\(q=1\)時滿足，此時\(a_{n}\)為非零常數(shù)列。3.如何根據(jù)數(shù)列的前幾項推測數(shù)列的通項公式？有哪些方法和注意事項？-答案：方法有觀察數(shù)字特征，如數(shù)字間的差、商規(guī)律；將數(shù)列與常見數(shù)列對比等。注意事項：僅根據(jù)前幾項推測的通項公式不唯一；要多驗證幾項；對于復(fù)雜數(shù)列可能需先對各項進(jìn)行變形處理。4.對于數(shù)列的前\(n\)項和\(S_{n}\)與通項\(a_{n}\)的關(guān)系\(a_{n}=\begin{cases}S_{1},n=1\\S_{n}-S_{n-1},n\geq2\end{cases}\)，在應(yīng)用中可能會遇到哪些問題？-答案：可能遇到\(n=1\)時\(a_{1}=S_{1}\)與\(n\geq2\)時\(a_{n}=S_{n}-S_{n-1}\)推出的表達(dá)式不統(tǒng)一，需單獨驗證\(n=1\)時是否滿足\(n\geq2\)的表達(dá)式；計算\(S_{n}-S_{n