春考數(shù)列題目及答案解析

以下是一份春考數(shù)列相關(guān)的試卷:單項選擇題:1.已知等差數(shù)列\(zhòng)(\{a_n\}\),\(a_3=5\),\(a_5=9\),則公差\(d\)為（）A.2B.1C.-2D.-1答案：A2.等比數(shù)列\(zhòng)(\{a_n\}\)中,\(a_2=2\),\(a_5=16\),則公比\(q\)為()A.2B.-2C.3D.-3答案：A3.數(shù)列\(zhòng)(\{a_n\}\)的通項公式\(a_n=2n-1\),則\(a_{10}\)為()A.19B.21C.29D.31答案：A4.等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項和\(S_n=2n^2+n\),則\(a_3\)為()A.11B.13C.15D.17答案：C5.等比數(shù)列\(zhòng)(\{a_n\}\)的首項\(a_1=1\),公比\(q=2\),則\(S_5\)為()A.31B.32C.63D.64答案：A6.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}=a_n+2\),\(a_1=1\),則\(a_5\)為()A.7B.9C.11D.13答案：B7.等差數(shù)列\(zhòng)(\{a_n\}\)中,若\(a_3+a_4=12\),\(a_6=7\),則公差\(d\)為()A.2B.3C.4D.5答案：A8.等比數(shù)列\(zhòng)(\{a_n\}\)中,\(a_3=4\),\(a_7=16\),則\(a_5\)為()A.8B.-8C.8或-8D.4答案：A9.數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項和\(S_n=n^2\),則\(a_n\)為()A.2n-1B.2nC.n^2-(n-1)^2D.n^2-(n+1)^2答案：A10.等差數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1+a_5=10\),\(a_4=7\),則公差\(d\)為()A.2B.3C.4D.5答案：A多項選擇題:1.等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項和\(S_n\),下列結(jié)論正確的是()A.若\(S_9>S_{10}\),則\(S_{10}>S_{11}\)B.若\(S_5=S_9\),則\(S_{14}=0\)C.若\(S_6>S_7\),則\(S_7>S_8\)D.若\(S_{10}>S_{11}\),則\(S_{11}>S_{12}\)答案:ABCD2.等比數(shù)列\(zhòng)(\{a_n\}\)的公比\(q\neq1\),下列結(jié)論正確的是()A.\(a_m\cdota_n=a_{m+n}\)B.\(\frac{a_m}{a_n}=q^{m-n}\)C.若\(m+n=p+q\),則\(a_m\cdota_n=a_p\cdota_q\)D.\(a_m+a_n\neqa_{m+n}\)答案:BCD3.已知數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}=2a_n\),\(a_1=1\),則()A.數(shù)列\(zhòng)(\{a_n\}\)是等比數(shù)列B.數(shù)列\(zhòng)(\{a_n\}\)的通項公式為\(a_n=2^{n-1}\)C.數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項和\(S_n=2^n-1\)D.數(shù)列\(zhòng)(\{a_n\}\)是遞增數(shù)列答案:ABCD4.等差數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=2\),\(d=3\),則()A.\(a_n=3n-1\)B.\(S_n=\frac{3}{2}n^2+\frac{1}{2}n\)C.\(a_{10}=29\)D.\(S_{10}=155\)答案:ABCD5.等比數(shù)列\(zhòng)(\{a_n\}\)中,\(a_2=4\),\(a_5=32\),則()A.公比\(q=2\)B.\(a_1=2\)C.\(a_n=2^n\)D.\(S_5=62\)答案:ABCD判斷題：1.等差數(shù)列\(zhòng)(\{a_n\}\)中,若\(m+n=p+q\),則\(a_m+a_n=a_p+a_q\)。()答案：對2.等比數(shù)列\(zhòng)(\{a_n\}\)中,若\(m+n=p+q\),則\(a_m\cdota_n=a_p\cdota_q\)。()答案：對3.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)。()答案：錯4.等比數(shù)列\(zhòng)(\{a_n\}\)的公比\(q\neq0\)。()答案：對5.數(shù)列\(zhòng)(\{a_n\}\)的通項公式\(a_n=2n-1\),則數(shù)列\(zhòng)(\{a_n\}\)是遞增數(shù)列。()答案：對6.等差數(shù)列\(zhòng)(\{a_n\}\)中,若\(a_1<0\),\(d>0\),則數(shù)列\(zhòng)(\{a_n\}\)一定是遞增數(shù)列。()答案：對7.等比數(shù)列\(zhòng)(\{a_n\}\)中,若\(a_1>0\),\(q>1\),則數(shù)列\(zhòng)(\{a_n\}\)一定是遞增數(shù)列。()答案：對8.數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項和\(S_n=n^2+1\),則\(a_n=2n-1\)。()答案：錯9.等差數(shù)列\(zhòng)(\{a_n\}\)中,若\(S_9=S_{18}\),則\(S_{27}=0\)。()答案：對10.等比數(shù)列\(zhòng)(\{a_n\}\)中,若\(a_1=1\),\(q=-1\),則\(S_{10}=0\)。()答案：對簡答題：1.已知等差數(shù)列\(zhòng)(\{a_n\}\),\(a_1=3\),\(d=2\),求\(a_{10}\)。答案:\(a_{10}=a_1+(10-1)d=3+9\times2=21\)。2.等比數(shù)列\(zhòng)(\{a_n\}\)中,\(a_2=6\),\(q=2\),求\(S_5\)。答案：\(a_1=\frac{a_2}{q}=\frac{6}{2}=3\),\(S_5=\frac{a_1(1-q^5)}{1-q}=\frac{3\times(1-2^5)}{1-2}=93\)。3.已知數(shù)列\(zhòng)(\{a_n\}\)的通項公式\(a_n=3n-2\),求\(S_{10}\)。答案：\(a_1=3\times1-2=1\),\(S_{10}=\frac{10\times(a_1+a_{10})}{2}=\frac{10\times(1+3\times10-2)}{2}=145\)。4.等差數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=1\),\(S_{10}=100\),求公差\(d\)。答案：由\(S_{10}=10a_1+\frac{10\times9}{2}d=100\),把\(a_1=1\)代入可得\(10+45d=100\),解得\(d=2\)。討論題：1.討論等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項和\(S_n\)的性質(zhì)。答案:等差數(shù)列前\(n\)項和\(S_n\)是關(guān)于\(n\)的二次函數(shù)形式(當(dāng)公差不為\(0\)時),且\(S_n\)的最值可通過對稱軸等方法來確定,還可根據(jù)公差的正負判斷數(shù)列的單調(diào)性等。2.討論等比數(shù)列\(zhòng)(\{a_n\}\)的公比\(q\)對數(shù)列性質(zhì)的影響。答案:公比\(q\)決定等比數(shù)列的單調(diào)性,當(dāng)\(q>1\)且\(a_1>0\)或\(0<q<1\)且\(a_1<0\)時數(shù)列遞增;當(dāng)\(q>1\)且\(a_1<0\)或\(0<q<1\)且\(a_1>0\)時數(shù)列遞減。同時\(q\neq0\),\(q\)還影響等比數(shù)列各項的比值關(guān)系等。3.已知一個數(shù)列的前\(n\)項和公式,如何判斷該數(shù)列是等差數(shù)列還是等比數(shù)列?答案：若前\(n\)項和公式是關(guān)于\(n\)的二次函數(shù)形式（不含常數(shù)項）則為等差數(shù)列；若前\(n\)項和公式是\(A(1-q^n)\div(1