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試卷第=page11頁,共=sectionpages33頁專題11數(shù)列多選題1.(2023秋·浙江·高三浙江省永康市第一中學(xué)校聯(lián)考期末)數(shù)列SKIPIF1<0的通項為SKIPIF1<0,它的前SKIPIF1<0項和為SKIPIF1<0,前SKIPIF1<0項積為SKIPIF1<0,則下列說法正確的是(

)A.?dāng)?shù)列SKIPIF1<0是遞減數(shù)列 B.當(dāng)SKIPIF1<0或者SKIPIF1<0時,SKIPIF1<0有最大值C.當(dāng)SKIPIF1<0或者SKIPIF1<0時,SKIPIF1<0有最大值 D.SKIPIF1<0和SKIPIF1<0都沒有最小值【答案】ABC【分析】根據(jù)數(shù)列的通項得出數(shù)列SKIPIF1<0是以SKIPIF1<0為首項,以SKIPIF1<0為公差的等差數(shù)列,然后根據(jù)等差數(shù)列的特征分別對每個選項進(jìn)行分析即可求解.【詳解】因?yàn)閿?shù)列SKIPIF1<0的通項為SKIPIF1<0,則SKIPIF1<0,所以數(shù)列SKIPIF1<0是以SKIPIF1<0為首項,以SKIPIF1<0為公差的等差數(shù)列,因?yàn)楣頢KIPIF1<0,所以數(shù)列SKIPIF1<0是遞減數(shù)列,故選項SKIPIF1<0正確;因?yàn)镾KIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0;當(dāng)SKIPIF1<0時,SKIPIF1<0,因?yàn)镾KIPIF1<0,所以當(dāng)SKIPIF1<0或者SKIPIF1<0時,SKIPIF1<0有最大值,故選項SKIPIF1<0正確;由SKIPIF1<0可知:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以當(dāng)SKIPIF1<0或者SKIPIF1<0時,SKIPIF1<0有最大值,故選項SKIPIF1<0正確;根據(jù)數(shù)列前30項為正數(shù),從第31項開始為負(fù)數(shù)可知:SKIPIF1<0無最小值,因?yàn)镾KIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,但零乘任何數(shù)仍得零,所以SKIPIF1<0有最小值SKIPIF1<0,故選項SKIPIF1<0錯誤,故選:SKIPIF1<0.2.(2023·廣東梅州·統(tǒng)考一模)設(shè)SKIPIF1<0是公差為SKIPIF1<0(SKIPIF1<0)的無窮等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和,則下列命題正確的是(

)A.若SKIPIF1<0,則SKIPIF1<0是數(shù)列SKIPIF1<0的最大項B.若數(shù)列SKIPIF1<0有最小項,則SKIPIF1<0C.若數(shù)列SKIPIF1<0是遞減數(shù)列,則對任意的:SKIPIF1<0,均有SKIPIF1<0D.若對任意的SKIPIF1<0,均有SKIPIF1<0,則數(shù)列SKIPIF1<0是遞增數(shù)列【答案】BD【分析】取特殊數(shù)列判斷A;由等差數(shù)列前SKIPIF1<0項和的函數(shù)特性判斷B;取特殊數(shù)列結(jié)合數(shù)列的單調(diào)性判斷C;討論數(shù)列SKIPIF1<0是遞減數(shù)列的情況,從而證明D.【詳解】對于A:取數(shù)列SKIPIF1<0為首項為4,公差為SKIPIF1<0的等差數(shù)列,SKIPIF1<0,故A錯誤;對于B:等差數(shù)列SKIPIF1<0中,公差SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是關(guān)于n的二次函數(shù).當(dāng)數(shù)列SKIPIF1<0有最小項,即SKIPIF1<0有最小值,SKIPIF1<0對應(yīng)的二次函數(shù)有最小值,對應(yīng)的函數(shù)圖象開口向上,SKIPIF1<0,B正確;對于C:取數(shù)列SKIPIF1<0為首項為1,公差為SKIPIF1<0的等差數(shù)列,SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0恒成立,此時數(shù)列SKIPIF1<0是遞減數(shù)列,而SKIPIF1<0,故C錯誤;對于D:若數(shù)列SKIPIF1<0是遞減數(shù)列,則SKIPIF1<0,一定存在實(shí)數(shù)SKIPIF1<0,當(dāng)SKIPIF1<0時,之后所有項都為負(fù)數(shù),不能保證對任意SKIPIF1<0,均有SKIPIF1<0.故若對任意SKIPIF1<0,均有SKIPIF1<0,有數(shù)列SKIPIF1<0是遞增數(shù)列,故D正確.故選:BD3.(2023春·浙江溫州·高三統(tǒng)考開學(xué)考試)已知正m邊形SKIPIF1<0,一質(zhì)點(diǎn)M從SKIPIF1<0點(diǎn)出發(fā),每一步移動均為等可能的到達(dá)與其相鄰兩個頂點(diǎn)之一.經(jīng)過n次移動,記質(zhì)點(diǎn)M又回到SKIPIF1<0點(diǎn)的方式數(shù)共有SKIPIF1<0種,且其概率為SKIPIF1<0,則下列說法正確的是(

)A.若SKIPIF1<0,則SKIPIF1<0 B.若SKIPIF1<0,則SKIPIF1<0C.若SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0 D.若SKIPIF1<0,則SKIPIF1<0【答案】BCD【分析】根據(jù)所給規(guī)則,直接判斷A,根據(jù)規(guī)則,分析變化規(guī)律,歸納得出結(jié)論判斷B,根據(jù)規(guī)則直接判斷C,列舉所有可能由古典概型求解判斷D.【詳解】對A,SKIPIF1<0時,如圖,經(jīng)3步從SKIPIF1<0回到SKIPIF1<0,僅有SKIPIF1<0,與SKIPIF1<0兩種,所以SKIPIF1<0,故A錯誤;對B,若SKIPIF1<0時,如圖,SKIPIF1<0與SKIPIF1<0,記從SKIPIF1<0出發(fā)經(jīng)過n步到SKIPIF1<0的方法數(shù)為SKIPIF1<0,則SKIPIF1<0(先走兩步回到SKIPIF1<0有2種,化歸為SKIPIF1<0,先走兩步到SKIPIF1<0有2種,化歸為SKIPIF1<0),所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,故B正確;對C,SKIPIF1<0時,顯然走奇數(shù)步無法回到SKIPIF1<0,故SKIPIF1<0,故C正確;對D,SKIPIF1<0時,走6步共有SKIPIF1<0種走法(每一步順時針或逆時針),SKIPIF1<0出發(fā)回到SKIPIF1<0有2種情形,①一個方向連續(xù)走6步,有2種;②2個方向各走3步,有SKIPIF1<0種,所以SKIPIF1<0,所以SKIPIF1<0,故D正確.故選:BCD4.(2023·吉林·東北師大附中??级#┮阎獢?shù)列SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的前SKIPIF1<0項的和為SKIPIF1<0,前SKIPIF1<0項的積為SKIPIF1<0,則下列結(jié)論正確的是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】BCD【分析】令SKIPIF1<0可求得SKIPIF1<0的值,推導(dǎo)出SKIPIF1<0,分析可知數(shù)列SKIPIF1<0中的奇數(shù)項和偶數(shù)項分別成以SKIPIF1<0為公比的等比數(shù)列,求出數(shù)列SKIPIF1<0的通項公式,逐項判斷可得出合適的選項.【詳解】數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時,則有SKIPIF1<0,可得SKIPIF1<0,當(dāng)SKIPIF1<0時,由SKIPIF1<0可得SKIPIF1<0,上述兩個等式相除可得SKIPIF1<0,B對;所以,數(shù)列SKIPIF1<0中的奇數(shù)項和偶數(shù)項分別成以SKIPIF1<0為公比的等比數(shù)列,當(dāng)SKIPIF1<0為奇數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0為偶數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,故對任意的SKIPIF1<0,SKIPIF1<0,所以,SKIPIF1<0,A錯;SKIPIF1<0,所以數(shù)列SKIPIF1<0為等比數(shù)列,且該數(shù)列的首項為SKIPIF1<0,公比為SKIPIF1<0,則SKIPIF1<0,C對;SKIPIF1<0,D對.故選:BCD.5.(2023秋·吉林遼源·高三校聯(lián)考期末)設(shè)SKIPIF1<0,SKIPIF1<0分別為等差數(shù)列SKIPIF1<0的公差與前n項和,若SKIPIF1<0,則下列論斷中正確的有(

)A.當(dāng)SKIPIF1<0時,SKIPIF1<0取最大值 B.當(dāng)SKIPIF1<0時,SKIPIF1<0C.當(dāng)SKIPIF1<0時,SKIPIF1<0 D.當(dāng)SKIPIF1<0時,SKIPIF1<0【答案】BCD【分析】根據(jù)等差數(shù)列的通項公式及前n項和公式,結(jié)合二次函數(shù)的性質(zhì)即可求解.【詳解】設(shè)等差數(shù)列SKIPIF1<0的首項為SKIPIF1<0,則由SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,當(dāng)SKIPIF1<0時,當(dāng)SKIPIF1<0時,SKIPIF1<0取最小值;當(dāng)SKIPIF1<0時,當(dāng)SKIPIF1<0時,SKIPIF1<0取最大值;故A錯誤;當(dāng)SKIPIF1<0時,SKIPIF1<0,故B正確;當(dāng)SKIPIF1<0時,SKIPIF1<0,故C正確;當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故D正確.故選:BCD.6.(2023秋·黑龍江哈爾濱·高三哈爾濱市第六中學(xué)校校考期末)數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,則下列結(jié)論正確的有(

)A.SKIPIF1<0B.?dāng)?shù)列SKIPIF1<0的和為SKIPIF1<0C.若數(shù)列SKIPIF1<0,則數(shù)列SKIPIF1<0D.?dāng)?shù)列SKIPIF1<0有最小項【答案】ABC【分析】逐項代入分析即可求解.【詳解】根據(jù)SKIPIF1<0,所以SKIPIF1<0為等差數(shù)列,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故選項A正確;SKIPIF1<0,所以數(shù)列SKIPIF1<0的和為SKIPIF1<0,故選項B正確;SKIPIF1<0,SKIPIF1<0,故選項C正確;令SKIPIF1<0,所以所以SKIPIF1<0,所以SKIPIF1<0,所以數(shù)列SKIPIF1<0沒有最小項,故選項D錯誤;故選:ABC.7.(2023·云南昆明·昆明一中??寄M預(yù)測)數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0,且SKIPIF1<0,則下列正確的是(

)A.SKIPIF1<0B.?dāng)?shù)列SKIPIF1<0的前n項和SKIPIF1<0C.?dāng)?shù)列SKIPIF1<0的前n項和SKIPIF1<0D.SKIPIF1<0【答案】BCD【分析】求得數(shù)列SKIPIF1<0的通項公式判斷選項A;求得數(shù)列SKIPIF1<0的前n項判斷選項B;求得數(shù)列SKIPIF1<0的前n項和,進(jìn)而判斷選項C;求得數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0進(jìn)而判斷選項D.【詳解】由SKIPIF1<0,有SKIPIF1<0,又SKIPIF1<0所以SKIPIF1<0是首項為SKIPIF1<0,公差為SKIPIF1<0的等差數(shù)列,則SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0,A錯誤;由SKIPIF1<0,可得SKIPIF1<0,解之得SKIPIF1<0又SKIPIF1<0時,SKIPIF1<0,則SKIPIF1<0,整理得SKIPIF1<0則數(shù)列SKIPIF1<0是首項為3公比為3的等比數(shù)列,則SKIPIF1<0,則數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0SKIPIF1<0,B正確;SKIPIF1<0,則數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0,C正確;設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,兩式相減得SKIPIF1<0整理得SKIPIF1<0,則當(dāng)SKIPIF1<0時,SKIPIF1<0,D正確.故選:BCD.8.(2023·云南昆明·高三昆明一中校考階段練習(xí))已知各項均為正數(shù)的數(shù)列SKIPIF1<0滿足:SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0是數(shù)列SKIPIF1<0的前n項和,則(

)A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】BCD【分析】A.將條件變形,利用求根公式,即可求解;B.根據(jù)通項公式求SKIPIF1<0;C.作除法,和1比較大小,即可判斷;D.利用通項公式求SKIPIF1<0,再構(gòu)造函數(shù)證明SKIPIF1<0,利用不等式變形,結(jié)合等差數(shù)列求和,即可證明.【詳解】A.SKIPIF1<0,變形為SKIPIF1<0,根據(jù)求根公式可知SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,故A錯誤;B.SKIPIF1<0,故B正確;C.SKIPIF1<0,SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0(SKIPIF1<0),故C正確;D.SKIPIF1<0SKIPIF1<0所以SKIPIF1<0SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增,當(dāng)SKIPIF1<0時,SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減,所以當(dāng)SKIPIF1<0時,函數(shù)SKIPIF1<0取得最大值0,所以SKIPIF1<0,即SKIPIF1<0,當(dāng)SKIPIF1<0時,等號成立,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0,故D正確.故選:BCD9.(2023春·安徽·高三合肥市第六中學(xué)校聯(lián)考開學(xué)考試)數(shù)列SKIPIF1<0滿足:SKIPIF1<0,SKIPIF1<0,則下列結(jié)論中正確的是(

)A.SKIPIF1<0 B.SKIPIF1<0,SKIPIF1<0C.SKIPIF1<0是等比數(shù)列 D.SKIPIF1<0,SKIPIF1<0【答案】ABD【分析】令SKIPIF1<0得出SKIPIF1<0,可判斷選項A;由已知構(gòu)造SKIPIF1<0與已知等式作差,可判斷選項B,C;數(shù)列SKIPIF1<0的首項為SKIPIF1<0,從第2項開始構(gòu)成等比數(shù)列,求和即可判斷選項D.【詳解】在SKIPIF1<0中,令SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.A正確.當(dāng)SKIPIF1<0時,將SKIPIF1<0與SKIPIF1<0,兩式相減得,SKIPIF1<0,即SKIPIF1<0.而SKIPIF1<0,所以B正確,C不正確.因?yàn)镾KIPIF1<0,SKIPIF1<0,所以D正確.故選:ABD.10.(2023·遼寧沈陽·統(tǒng)考一模)SKIPIF1<0是各項均為正數(shù)的等差數(shù)列,其公差SKIPIF1<0,SKIPIF1<0是等比數(shù)列,若SKIPIF1<0,SKIPIF1<0,則(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】AD【分析】對于函數(shù)SKIPIF1<0的零點(diǎn)為SKIPIF1<0,分類討論結(jié)合導(dǎo)數(shù)分析可得當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,即可判斷結(jié)果.【詳解】由題意可設(shè):等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0,由SKIPIF1<0,SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,構(gòu)建SKIPIF1<0,即SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0為函數(shù)SKIPIF1<0的零點(diǎn),當(dāng)SKIPIF1<0時,則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,故SKIPIF1<0在SKIPIF1<0內(nèi)至多有一個零點(diǎn),不合題意;當(dāng)SKIPIF1<0時,則SKIPIF1<0,∵SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,故SKIPIF1<0在SKIPIF1<0上至多有一個零點(diǎn),當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)無零點(diǎn),則SKIPIF1<0在SKIPIF1<0上恒成立,故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,則SKIPIF1<0在SKIPIF1<0內(nèi)至多有一個零點(diǎn),不合題意;當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)有零點(diǎn),設(shè)為SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,且SKIPIF1<0,可得當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,故SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,A、D正確,B、C錯誤.故選:AD.11.(2023·遼寧·校聯(lián)考一模)設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和是SKIPIF1<0,若SKIPIF1<0,則(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】BC【分析】設(shè)等差數(shù)列公差為d,由題目條件,可得SKIPIF1<0,由此可得各選項正誤.【詳解】設(shè)等差數(shù)列公差為d,則由題目條件有:SKIPIF1<0.A選項,SKIPIF1<0,故A錯誤;B選項,SKIPIF1<0,故B正確;C選項,SKIPIF1<0,故C正確;D選項,注意到SKIPIF1<0,SKIPIF1<0,又由SKIPIF1<0知SKIPIF1<0為單調(diào)遞減數(shù)列,則SKIPIF1<0,故D錯誤.故選:BC.12.(2023秋·江蘇·高三南京師大附中校聯(lián)考期末)已知數(shù)列SKIPIF1<0為等差數(shù)列,首項為1,公差為2,數(shù)列SKIPIF1<0為等比數(shù)列,首項為1,公比為2,設(shè)SKIPIF1<0,SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項和,則當(dāng)SKIPIF1<0時,SKIPIF1<0的取值可以是下面選項中的(

)A.8 B.9 C.10 D.11【答案】AB【分析】由已知分別得到等差數(shù)列與等比數(shù)列的通項公式,求得數(shù)列SKIPIF1<0的通項公式,利用數(shù)列的分組求和法可得數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0,驗(yàn)證得答案.【詳解】依題意得:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則數(shù)列SKIPIF1<0為遞增數(shù)列,其前SKIPIF1<0項和SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0的取值可以是8,9,故選:AB.13.(2023秋·河北衡水·高三河北衡水中學(xué)??计谀┤魯?shù)列SKIPIF1<0有SKIPIF1<0,SKIPIF1<0為SKIPIF1<0前n項積,SKIPIF1<0有SKIPIF1<0,則(

)A.SKIPIF1<0為等差數(shù)列(SKIPIF1<0) B.可能SKIPIF1<0C.SKIPIF1<0為等差數(shù)列 D.SKIPIF1<0第n項可能與n無關(guān)【答案】BD【分析】結(jié)合遞推式SKIPIF1<0,取SKIPIF1<0,求SKIPIF1<0的通項公式判斷選項A錯誤,求SKIPIF1<0判斷B,由遞推式SKIPIF1<0,取SKIPIF1<0,判斷C,求數(shù)列SKIPIF1<0的通項公式判斷D.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以當(dāng)SKIPIF1<0時,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0不存在,A錯誤;因?yàn)镾KIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以可能SKIPIF1<0,B正確;因?yàn)镾KIPIF1<0,取SKIPIF1<0,則SKIPIF1<0,此時SKIPIF1<0不存在,C錯誤;D正確;故選:BD.14.(2023秋·山東濱州·高三統(tǒng)考期末)已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則下列結(jié)論正確的是(

)A.SKIPIF1<0 B.SKIPIF1<0是等比數(shù)列C.SKIPIF1<0是單調(diào)遞增數(shù)列 D.SKIPIF1<0【答案】AC【分析】由已知得出SKIPIF1<0,可判斷A選項的正誤;利用等比數(shù)列的定義可判斷B選項的正誤;利用數(shù)列的單調(diào)性可判斷C選項的正誤;利用作差法可判斷D選項的正誤.【詳解】對于A選項,由SKIPIF1<0得SKIPIF1<0,故SKIPIF1<0,A正確;對于B選項,將SKIPIF1<0,SKIPIF1<0兩式相減得SKIPIF1<0,即SKIPIF1<0SKIPIF1<0,又令SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0從第二項開始成等比數(shù)列,公比為SKIPIF1<0,故SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0,所以,SKIPIF1<0,故B選項錯誤;對于C選項,因?yàn)镾KIPIF1<0.當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0.所以,SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0,而SKIPIF1<0,所以數(shù)列SKIPIF1<0單調(diào)遞增,C選項正確;對于D選項,當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0顯然成立,故SKIPIF1<0恒成立,D選項錯誤.故選:AC.15.(2023秋·山東德州·高三統(tǒng)考期末)已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0則(

)A.SKIPIF1<0 B.SKIPIF1<0C.?dāng)?shù)列SKIPIF1<0為等差數(shù)列 D.SKIPIF1<0為奇數(shù)時,SKIPIF1<0【答案】ABD【分析】利用并項求和法可判斷AD選項;利用等差數(shù)列的定義可判斷BC選項.【詳解】對于A選項,SKIPIF1<0,A對;對于B選項,因?yàn)镾KIPIF1<0,則SKIPIF1<0,對任意的SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,上述兩個等式作差可得SKIPIF1<0,所以,數(shù)列SKIPIF1<0中的奇數(shù)項成以SKIPIF1<0為首項,公差為SKIPIF1<0的等差數(shù)列,數(shù)列SKIPIF1<0中的偶數(shù)項成以SKIPIF1<0為首項,公差為SKIPIF1<0的等差數(shù)列,當(dāng)SKIPIF1<0為奇數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0為偶數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,綜上所述,SKIPIF1<0,B對;對于C選項,SKIPIF1<0,故數(shù)列SKIPIF1<0不是等差數(shù)列,C錯;對于D選項,當(dāng)SKIPIF1<0為奇數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0SKIPIF1<0SKIPIF1<0,D對.故選:ABD.16.(2023秋·湖北武漢·高三統(tǒng)考期末)等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,前SKIPIF1<0項的積SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,則下列選項中成立的是(

)A.對任意正整數(shù)SKIPIF1<0,SKIPIF1<0 B.SKIPIF1<0C.?dāng)?shù)列SKIPIF1<0一定是等比數(shù)列 D.SKIPIF1<0【答案】ABC【分析】設(shè)公比為SKIPIF1<0,首項為SKIPIF1<0,依題意可得SKIPIF1<0,SKIPIF1<0,即可得到SKIPIF1<0,從而判斷數(shù)列的單調(diào)性,即可判斷BD,再根據(jù)等比數(shù)列前SKIPIF1<0項和公式及等比數(shù)列的定義判斷C,最后根據(jù)等比中項及作差法判斷A.【詳解】解:設(shè)公比為SKIPIF1<0,首項為SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以數(shù)列SKIPIF1<0是各項為正數(shù)的等比數(shù)列且SKIPIF1<0,所以數(shù)列SKIPIF1<0單調(diào)遞減,則SKIPIF1<0,故B正確,D錯誤;所以SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以數(shù)列SKIPIF1<0是以SKIPIF1<0為首項,SKIPIF1<0為公比的等比數(shù)列,故C正確;因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,故A正確;故選:ABC17.(2023·湖北·校聯(lián)考模擬預(yù)測)數(shù)列SKIPIF1<0各項均為正數(shù),其前n項和SKIPIF1<0,且滿足SKIPIF1<0,下列四個結(jié)論中正確的是(

)A.SKIPIF1<0為等比數(shù)列 B.SKIPIF1<0為遞減數(shù)列C.SKIPIF1<0中存在大于3的項 D.SKIPIF1<0中存在小于SKIPIF1<0的項【答案】BD【分析】對于A:假設(shè)數(shù)列SKIPIF1<0為等比數(shù)列,設(shè)其公比為q,求出SKIPIF1<0,不合乎題意;對于B:求出SKIPIF1<0,得到SKIPIF1<0,即可證明;對于C:先求出SKIPIF1<0,由數(shù)列SKIPIF1<0為遞減數(shù)列,即可判斷;對于D:利用單調(diào)性證明.【詳解】對于A:假設(shè)數(shù)列SKIPIF1<0為等比數(shù)列,設(shè)其公比為q,則SKIPIF1<0,即SKIPIF1<0,所以,SKIPIF1<0,可得SKIPIF1<0,解得SKIPIF1<0,不合乎題意,故數(shù)列SKIPIF1<0不是等比數(shù)列,故A錯;對于B:當(dāng)SKIPIF1<0時,SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,可得SKIPIF1<0,所以數(shù)列SKIPIF1<0為遞減數(shù)列,故B對;對于C:由題意可知,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,可得SKIPIF1<0;由B知數(shù)列SKIPIF1<0為遞減數(shù)列,故C錯;對于D:因?yàn)閿?shù)列SKIPIF1<0各項均為正數(shù),其前n項和SKIPIF1<0,所以隨著n的增大,SKIPIF1<0遞增.而SKIPIF1<0恒成立,所以SKIPIF1<0遞減,且SKIPIF1<0,所以SKIPIF1<0中必存在小于SKIPIF1<0的項故選:BD.18.(2023秋·江蘇南京·高三南京市第一中學(xué)??计谀┮阎獢?shù)列SKIPIF1<0SKIPIF1<0的項數(shù)均為SKIPIF1<0(SKIPIF1<0為確定的正整數(shù),且SKIPIF1<0),若SKIPIF1<0,SKIPIF1<0,則(

)A.SKIPIF1<0中可能有SKIPIF1<0項為1 B.SKIPIF1<0中至多有SKIPIF1<0項為1C.SKIPIF1<0可能是以SKIPIF1<0為公比的等比數(shù)列 D.SKIPIF1<0可能是以2為公比的等比數(shù)列【答案】AC【分析】利用SKIPIF1<0求出數(shù)列SKIPIF1<0SKIPIF1<0,再根據(jù)SKIPIF1<0的取值判斷即可.【詳解】由題意可得SKIPIF1<0①,SKIPIF1<0②,①-②得SKIPIF1<0,同理可得SKIPIF1<0,所以數(shù)列SKIPIF1<0SKIPIF1<0中僅有1項為1,因?yàn)镾KIPIF1<0,所以B錯誤;當(dāng)SKIPIF1<0時,A正確;SKIPIF1<0,所以當(dāng)SKIPIF1<0時,SKIPIF1<0是以SKIPIF1<0為公比的等比數(shù)列,C正確,D錯誤;故選:AC19.(2023春·廣東惠州·高三校考階段練習(xí))斐波那契數(shù)列又稱黃金分割數(shù)列,因數(shù)學(xué)家列昂納多·斐波那契以兔子繁殖為例子而引入,故又稱為“兔子數(shù)列”.斐波那契數(shù)列用遞推的方式可如下定義:用SKIPIF1<0表示斐波那契數(shù)列的第n項,則數(shù)列SKIPIF1<0滿足:SKIPIF1<0,SKIPIF1<0,記SKIPIF1<0,則下列結(jié)論正確的是(

)A.?dāng)?shù)列SKIPIF1<0是遞增數(shù)列 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】BCD【分析】由數(shù)列的遞推公式可判斷A,B;利用累加法計算可判斷選項C,D.【詳解】對A,由SKIPIF1<0知,SKIPIF1<0的前10項依次為:1,1,2,3,5,8,13,21,34,55,其中,第一二項相等,不滿足遞增性,故A錯誤;對B,根據(jù)遞推公式SKIPIF1<0,得SKIPIF1<0,故B正確;對C,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,……,SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,故C正確;對D,由遞推式,得SKIPIF1<0,SKIPIF1<0,…,SKIPIF1<0,累加得SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,故D正確;故選:BCD.20.(2023春·浙江·高三校聯(lián)考開學(xué)考試)已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項和,則(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】AD【分析】對于A,證明數(shù)列SKIPIF1<0單調(diào)遞減即得解;對于B,證明SKIPIF1<0即得解;對于C,隨著SKIPIF1<0減小,從而SKIPIF1<0增大.即得解;對于D,證明SKIPIF1<0即得解.【詳解】對于A:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0單調(diào)遞增,SKIPIF1<0在SKIPIF1<0單調(diào)遞減,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時,SKIPIF1<0若SKIPIF1<0,又因?yàn)镾KIPIF1<0則SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0,又因?yàn)镾KIPIF1<0所以SKIPIF1<0所以SKIPIF1<0,設(shè)SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0時,SKIPIF1<0SKIPIF1<0單調(diào)遞增.所以SKIPIF1<0所以SKIPIF1<0所以SKIPIF1<0由SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0因?yàn)镾KIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0,同理得SKIPIF1<0,SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時,SKIPIF1<0;所以SKIPIF1<0,所以數(shù)列SKIPIF1<0單調(diào)遞減.則SKIPIF1<0,所以選項A正確.對于B:由前面得SKIPIF1<0.下面證明SKIPIF1<0.只需證明SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,∴SKIPIF1<0成立.所以SKIPIF1<0,所以SKIPIF1<0,所以選項B錯誤;對于C:SKIPIF1<0,設(shè)SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0.所以函數(shù)SKIPIF1<0單調(diào)遞減,所以隨著SKIPIF1<0減小,從而SKIPIF1<0增大.所以SKIPIF1<0,即SKIPIF1<0.所以C錯誤.對于D:一般地,證明:SKIPIF1<0.只需證明SKIPIF1<0.SKIPIF1<0.令SKIPIF1<0,則SKIPIF1<0,∴SKIPIF1<0成立.所以SKIPIF1<0,所以SKIPIF1<0.所以D正確.故選:SKIPIF1<0.21.(2023·浙江·校聯(lián)考模擬預(yù)測)數(shù)列SKIPIF1<0定義如下:SKIPIF1<0,SKIPIF1<0,若對于任意SKIPIF1<0,數(shù)列的前SKIPIF1<0項已定義,則對于SKIPIF1<0,定義SKIPIF1<0,SKIPIF1<0為其前n項和,則下列結(jié)論正確的是(

)A.?dāng)?shù)列SKIPIF1<0的第SKIPIF1<0項為SKIPIF1<0 B.?dāng)?shù)列SKIPIF1<0的第2023項為SKIPIF1<0C.?dāng)?shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0 D.SKIPIF1<0【答案】ACD【分析】由數(shù)列的定義,對通項和前n項和的性質(zhì)進(jìn)行討論,驗(yàn)證選項是否正確.【詳解】SKIPIF1<0SKIPIF1<0…,SKIPIF1<0,故A選項正確;SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故B選項錯誤;SKIPIF1<0,SKIPIF1<0,…,當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0,故C選項正確;當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0,故D選項正確;故選:ACD.【點(diǎn)睛】方法點(diǎn)睛:解決新定義問題,首先考查對定義的理解。其次是考查滿足新定義的數(shù)列的簡單應(yīng)用,如在某些條件下,滿足新定義的數(shù)列有某些新的性質(zhì),這也是在新環(huán)境下研究“舊”性質(zhì),此時需要結(jié)合新數(shù)列的新性質(zhì),探究“舊”性質(zhì).第三是考查綜合分析能力,主要是將新性質(zhì)有機(jī)地應(yīng)用在“舊”性質(zhì)上,創(chuàng)造性地證明更新的性質(zhì).22.(2023春·浙江寧波·高三校聯(lián)考階段練習(xí))數(shù)列SKIPIF1<0前SKIPIF1<0項和為SKIPIF1<0,若SKIPIF1<0,且SKIPIF1<0,則以下結(jié)論正確的有(

)A.SKIPIF1<0B.?dāng)?shù)列SKIPIF1<0為遞增數(shù)列C.?dāng)?shù)列SKIPIF1<0為等差數(shù)列D.SKIPIF1<0的最大值為SKIPIF1<0【答案】BCD【分析】對A:取特值,結(jié)合SKIPIF1<0,運(yùn)算求解即可;對B:根據(jù)題意可得SKIPIF1<0,結(jié)合數(shù)列單調(diào)性分析判斷;對C:可得SKIPIF1<0,作差即可得結(jié)果;對D:利用累加法求得SKIPIF1<0,整理可得SKIPIF1<0,結(jié)合對勾函數(shù)的單調(diào)性分析運(yùn)算.【詳解】由SKIPIF1<0,可得:對A:令SKIPIF1<0可得:SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0可得:SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0,由SKIPIF1<0,解得SKIPIF1<0,A錯誤;對B:對SKIPIF1<0,則SKIPIF1<0,故數(shù)列SKIPIF1<0為遞增數(shù)列,B正確;對C:當(dāng)SKIPIF1<0時,可得SKIPIF1<0,則SKIPIF1<0,故數(shù)列SKIPIF1<0為等差數(shù)列,C正確;對D:∵SKIPIF1<0,則SKIPIF1<0SKIPIF1<0,且SKIPIF1<0,故SKIPIF1<0且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,且SKIPIF1<0,可得SKIPIF1<0,對SKIPIF1<0恒成立,故當(dāng)SKIPIF1<0時,SKIPIF1<0取最大值SKIPIF1<0,D正確.故選:BCD.【點(diǎn)睛】關(guān)鍵點(diǎn)睛:解決數(shù)列與函數(shù)綜合問題的注意點(diǎn)(1)數(shù)列是一類特殊的函數(shù),其定義域是正整數(shù)集,而不是某個區(qū)間上的連續(xù)實(shí)數(shù),所以它的圖象是一群孤立的點(diǎn).(2)轉(zhuǎn)化以函數(shù)為背景的條件時,應(yīng)注意題中的限制條件,如函數(shù)的定義域,這往往是非常容易忽視的問題.(3)利用函數(shù)的方法研究數(shù)列中相關(guān)問題時,應(yīng)準(zhǔn)確構(gòu)造函數(shù),注意數(shù)列中相關(guān)限制條件的轉(zhuǎn)化.23.(2023·云南昆明·安寧市第一中學(xué)??寄M預(yù)測)古希臘畢達(dá)哥拉斯學(xué)派的數(shù)學(xué)家用沙粒和小石子來研究數(shù),他們根據(jù)沙?;蛐∈铀帕械男螤睿褦?shù)分成許多類,如圖中第一行圖形中黑色小點(diǎn)個數(shù):1,3,6,10,…稱為三角形數(shù),第二行圖形中黑色小點(diǎn)個數(shù):1,4,9,16,…稱為正方形數(shù),記三角形數(shù)構(gòu)成數(shù)列SKIPIF1<0,正方形數(shù)構(gòu)成數(shù)列SKIPIF1<0,則下列說法正確的是(

)A.SKIPIF1<0B.1225既是三角形數(shù),又是正方形數(shù)C.SKIPIF1<0D.SKIPIF1<0,SKIPIF1<0,總存在SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立【答案】BCD【分析】根據(jù)給定信息,求出數(shù)列SKIPIF1<0、SKIPIF1<0的通項,再逐一分析各個選項即可判斷作答.【詳解】依題意,數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,于是得SKIPIF1<0,SKIPIF1<0滿足上式,數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,于是得SK

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