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試卷第=page11頁(yè),共=sectionpages33頁(yè)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(xiàng)為SKIPIF1<0,它的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,前SKIPIF1<0項(xiàng)積為SKIPIF1<0,則下列說(shuō)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(shí),SKIPIF1<0有最大值C.當(dāng)SKIPIF1<0或者SKIPIF1<0時(shí),SKIPIF1<0有最大值 D.SKIPIF1<0和SKIPIF1<0都沒(méi)有最小值【答案】ABC【分析】根據(jù)數(shù)列的通項(xiàng)得出數(shù)列SKIPIF1<0是以SKIPIF1<0為首項(xiàng),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(duì)每個(gè)選項(xiàng)進(jìn)行分析即可求解.【詳解】因?yàn)閿?shù)列SKIPIF1<0的通項(xiàng)為SKIPIF1<0,則SKIPIF1<0,所以數(shù)列SKIPIF1<0是以SKIPIF1<0為首項(xiàng),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(xiàng)SKIPIF1<0正確;因?yàn)镾KIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,因?yàn)镾KIPIF1<0,所以當(dāng)SKIPIF1<0或者SKIPIF1<0時(shí),SKIPIF1<0有最大值,故選項(xiàng)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(shí),SKIPIF1<0有最大值,故選項(xiàng)SKIPIF1<0正確;根據(jù)數(shù)列前30項(xiàng)為正數(shù),從第31項(xiàng)開始為負(fù)數(shù)可知:SKIPIF1<0無(wú)最小值,因?yàn)镾KIPIF1<0,當(dāng)SKIPIF1<0時(shí),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(xiàng)SKIPIF1<0錯(cuò)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(wú)窮等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)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(xiàng)B.若數(shù)列SKIPIF1<0有最小項(xiàng),則SKIPIF1<0C.若數(shù)列SKIPIF1<0是遞減數(shù)列,則對(duì)任意的:SKIPIF1<0,均有SKIPIF1<0D.若對(duì)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(xiàng)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(duì)于A:取數(shù)列SKIPIF1<0為首項(xiàng)為4,公差為SKIPIF1<0的等差數(shù)列,SKIPIF1<0,故A錯(cuò)誤;對(duì)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(xiàng),即SKIPIF1<0有最小值,SKIPIF1<0對(duì)應(yīng)的二次函數(shù)有最小值,對(duì)應(yīng)的函數(shù)圖象開口向上,SKIPIF1<0,B正確;對(duì)于C:取數(shù)列SKIPIF1<0為首項(xiàng)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(shí)數(shù)列SKIPIF1<0是遞減數(shù)列,而SKIPIF1<0,故C錯(cuò)誤;對(duì)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(shí),之后所有項(xiàng)都為負(fù)數(shù),不能保證對(duì)任意SKIPIF1<0,均有SKIPIF1<0.故若對(duì)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(dòng)均為等可能的到達(dá)與其相鄰兩個(gè)頂點(diǎn)之一.經(jīng)過(guò)n次移動(dòng),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(shuō)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(duì)A,SKIPIF1<0時(shí),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(cuò)誤;對(duì)B,若SKIPIF1<0時(shí),如圖,SKIPIF1<0與SKIPIF1<0,記從SKIPIF1<0出發(fā)經(jīng)過(guò)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(duì)C,SKIPIF1<0時(shí),顯然走奇數(shù)步無(wú)法回到SKIPIF1<0,故SKIPIF1<0,故C正確;對(duì)D,SKIPIF1<0時(shí),走6步共有SKIPIF1<0種走法(每一步順時(shí)針或逆時(shí)針),SKIPIF1<0出發(fā)回到SKIPIF1<0有2種情形,①一個(gè)方向連續(xù)走6步,有2種;②2個(gè)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(xiàng)的和為SKIPIF1<0,前SKIPIF1<0項(xiàng)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(xiàng)和偶數(shù)項(xiàng)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(xiàng)公式,逐項(xiàng)判斷可得出合適的選項(xiàng).【詳解】數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則有SKIPIF1<0,可得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),由SKIPIF1<0可得SKIPIF1<0,上述兩個(gè)等式相除可得SKIPIF1<0,B對(duì);所以,數(shù)列SKIPIF1<0中的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)分別成以SKIPIF1<0為公比的等比數(shù)列,當(dāng)SKIPIF1<0為奇數(shù)時(shí),設(shè)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0為偶數(shù)時(shí),設(shè)SKIPIF1<0,則SKIPIF1<0,故對(duì)任意的SKIPIF1<0,SKIPIF1<0,所以,SKIPIF1<0,A錯(cuò);SKIPIF1<0,所以數(shù)列SKIPIF1<0為等比數(shù)列,且該數(shù)列的首項(xiàng)為SKIPIF1<0,公比為SKIPIF1<0,則SKIPIF1<0,C對(duì);SKIPIF1<0,D對(duì).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(xiàng)和,若SKIPIF1<0,則下列論斷中正確的有(

)A.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取最大值 B.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0C.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0 D.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0【答案】BCD【分析】根據(jù)等差數(shù)列的通項(xiàng)公式及前n項(xiàng)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(xiàng)為SKIPIF1<0,則由SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取最小值;當(dāng)SKIPIF1<0時(shí),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取最大值;故A錯(cuò)誤;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故B正確;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故C正確;當(dāng)SKIPIF1<0時(shí),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(xiàng)【答案】ABC【分析】逐項(xiàng)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(xiàng)A正確;SKIPIF1<0,所以數(shù)列SKIPIF1<0的和為SKIPIF1<0,故選項(xiàng)B正確;SKIPIF1<0,SKIPIF1<0,故選項(xiàng)C正確;令SKIPIF1<0,所以所以SKIPIF1<0,所以SKIPIF1<0,所以數(shù)列SKIPIF1<0沒(méi)有最小項(xiàng),故選項(xiàng)D錯(cuò)誤;故選:ABC.7.(2023·云南昆明·昆明一中校考模擬預(yù)測(cè))數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,數(shù)列SKIPIF1<0的前n項(xiàng)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(xiàng)和SKIPIF1<0C.?dāng)?shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0D.SKIPIF1<0【答案】BCD【分析】求得數(shù)列SKIPIF1<0的通項(xiàng)公式判斷選項(xiàng)A;求得數(shù)列SKIPIF1<0的前n項(xiàng)判斷選項(xiàng)B;求得數(shù)列SKIPIF1<0的前n項(xiàng)和,進(jìn)而判斷選項(xiàng)C;求得數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0進(jìn)而判斷選項(xiàng)D.【詳解】由SKIPIF1<0,有SKIPIF1<0,又SKIPIF1<0所以SKIPIF1<0是首項(xiàng)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(cuò)誤;由SKIPIF1<0,可得SKIPIF1<0,解之得SKIPIF1<0又SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,整理得SKIPIF1<0則數(shù)列SKIPIF1<0是首項(xiàng)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(xiàng)和SKIPIF1<0SKIPIF1<0,B正確;SKIPIF1<0,則數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0,C正確;設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)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(shí),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(xiàng)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(xiàng)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(xiàng)公式求SKIPIF1<0;C.作除法,和1比較大小,即可判斷;D.利用通項(xiàng)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(cuò)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(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減,所以當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0取得最大值0,所以SKIPIF1<0,即SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),等號(hào)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(xiàng)A;由已知構(gòu)造SKIPIF1<0與已知等式作差,可判斷選項(xiàng)B,C;數(shù)列SKIPIF1<0的首項(xiàng)為SKIPIF1<0,從第2項(xiàng)開始構(gòu)成等比數(shù)列,求和即可判斷選項(xiàng)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(shí),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(yáng)·統(tǒng)考一模)SKIPIF1<0是各項(xiàng)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(duì)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(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),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(shí),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(gè)零點(diǎn),不合題意;當(dāng)SKIPIF1<0時(shí),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(gè)零點(diǎn),當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)無(wú)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(gè)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(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),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(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,A、D正確,B、C錯(cuò)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(xiàng)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(xiàng)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(xiàng),SKIPIF1<0,故A錯(cuò)誤;B選項(xiàng),SKIPIF1<0,故B正確;C選項(xiàng),SKIPIF1<0,故C正確;D選項(xiàng),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(cuò)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(xiàng)為1,公差為2,數(shù)列SKIPIF1<0為等比數(shù)列,首項(xiàng)為1,公比為2,設(shè)SKIPIF1<0,SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,則當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的取值可以是下面選項(xiàng)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(xiàng)公式,求得數(shù)列SKIPIF1<0的通項(xiàng)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(xiàng)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(xiàng)和SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),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(xiàng)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(xiàng)可能與n無(wú)關(guān)【答案】BD【分析】結(jié)合遞推式SKIPIF1<0,取SKIPIF1<0,求SKIPIF1<0的通項(xiàng)公式判斷選項(xiàng)A錯(cuò)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(xiàng)公式判斷D.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0不存在,A錯(cuò)誤;因?yàn)镾KIPIF1<0時(shí),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(shí)SKIPIF1<0不存在,C錯(cuò)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(xiàng)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(xiàng)的正誤;利用等比數(shù)列的定義可判斷B選項(xiàng)的正誤;利用數(shù)列的單調(diào)性可判斷C選項(xiàng)的正誤;利用作差法可判斷D選項(xiàng)的正誤.【詳解】對(duì)于A選項(xiàng),由SKIPIF1<0得SKIPIF1<0,故SKIPIF1<0,A正確;對(duì)于B選項(xiàng),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(xiàng)開始成等比數(shù)列,公比為SKIPIF1<0,故SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0,所以,SKIPIF1<0,故B選項(xiàng)錯(cuò)誤;對(duì)于C選項(xiàng),因?yàn)镾KIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0.所以,SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0,而SKIPIF1<0,所以數(shù)列SKIPIF1<0單調(diào)遞增,C選項(xiàng)正確;對(duì)于D選項(xiàng),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0顯然成立,故SKIPIF1<0恒成立,D選項(xiàng)錯(cuò)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(xiàng)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(shí),SKIPIF1<0【答案】ABD【分析】利用并項(xiàng)求和法可判斷AD選項(xiàng);利用等差數(shù)列的定義可判斷BC選項(xiàng).【詳解】對(duì)于A選項(xiàng),SKIPIF1<0,A對(duì);對(duì)于B選項(xiàng),因?yàn)镾KIPIF1<0,則SKIPIF1<0,對(duì)任意的SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,上述兩個(gè)等式作差可得SKIPIF1<0,所以,數(shù)列SKIPIF1<0中的奇數(shù)項(xiàng)成以SKIPIF1<0為首項(xiàng),公差為SKIPIF1<0的等差數(shù)列,數(shù)列SKIPIF1<0中的偶數(shù)項(xiàng)成以SKIPIF1<0為首項(xiàng),公差為SKIPIF1<0的等差數(shù)列,當(dāng)SKIPIF1<0為奇數(shù)時(shí),設(shè)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0為偶數(shù)時(shí),設(shè)SKIPIF1<0,則SKIPIF1<0,綜上所述,SKIPIF1<0,B對(duì);對(duì)于C選項(xiàng),SKIPIF1<0,故數(shù)列SKIPIF1<0不是等差數(shù)列,C錯(cuò);對(duì)于D選項(xiàng),當(dāng)SKIPIF1<0為奇數(shù)時(shí),設(shè)SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0SKIPIF1<0SKIPIF1<0,D對(duì).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(xiàng)和為SKIPIF1<0,前SKIPIF1<0項(xiàng)的積SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,則下列選項(xiàng)中成立的是(

)A.對(duì)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(xiàng)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(xiàng)和公式及等比數(shù)列的定義判斷C,最后根據(jù)等比中項(xiàng)及作差法判斷A.【詳解】解:設(shè)公比為SKIPIF1<0,首項(xiàng)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(xiàng)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(cuò)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(xiàng),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(cè))數(shù)列SKIPIF1<0各項(xiàng)均為正數(shù),其前n項(xiàng)和SKIPIF1<0,且滿足SKIPIF1<0,下列四個(gè)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(xiàng) D.SKIPIF1<0中存在小于SKIPIF1<0的項(xiàng)【答案】BD【分析】對(duì)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(duì)于B:求出SKIPIF1<0,得到SKIPIF1<0,即可證明;對(duì)于C:先求出SKIPIF1<0,由數(shù)列SKIPIF1<0為遞減數(shù)列,即可判斷;對(duì)于D:利用單調(diào)性證明.【詳解】對(duì)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(cuò);對(duì)于B:當(dāng)SKIPIF1<0時(shí),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(duì);對(duì)于C:由題意可知,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,可得SKIPIF1<0;由B知數(shù)列SKIPIF1<0為遞減數(shù)列,故C錯(cuò);對(duì)于D:因?yàn)閿?shù)列SKIPIF1<0各項(xiàng)均為正數(shù),其前n項(xiàng)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(xiàng)故選:BD.18.(2023秋·江蘇南京·高三南京市第一中學(xué)校考期末)已知數(shù)列SKIPIF1<0SKIPIF1<0的項(xiàng)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(xiàng)為1 B.SKIPIF1<0中至多有SKIPIF1<0項(xiàng)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(xiàng)為1,因?yàn)镾KIPIF1<0,所以B錯(cuò)誤;當(dāng)SKIPIF1<0時(shí),A正確;SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0是以SKIPIF1<0為公比的等比數(shù)列,C正確,D錯(cuò)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(xiàng),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(jì)算可判斷選項(xiàng)C,D.【詳解】對(duì)A,由SKIPIF1<0知,SKIPIF1<0的前10項(xiàng)依次為:1,1,2,3,5,8,13,21,34,55,其中,第一二項(xiàng)相等,不滿足遞增性,故A錯(cuò)誤;對(duì)B,根據(jù)遞推公式SKIPIF1<0,得SKIPIF1<0,故B正確;對(duì)C,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,……,SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,故C正確;對(duì)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(xiàng)和,則(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】AD【分析】對(duì)于A,證明數(shù)列SKIPIF1<0單調(diào)遞減即得解;對(duì)于B,證明SKIPIF1<0即得解;對(duì)于C,隨著SKIPIF1<0減小,從而SKIPIF1<0增大.即得解;對(duì)于D,證明SKIPIF1<0即得解.【詳解】對(duì)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(shí),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(shí),SKIPIF1<0SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0時(shí),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(shí),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(shí),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(xiàng)A正確.對(duì)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(xiàng)B錯(cuò)誤;對(duì)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(cuò)誤.對(duì)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(cè))數(shù)列SKIPIF1<0定義如下:SKIPIF1<0,SKIPIF1<0,若對(duì)于任意SKIPIF1<0,數(shù)列的前SKIPIF1<0項(xiàng)已定義,則對(duì)于SKIPIF1<0,定義SKIPIF1<0,SKIPIF1<0為其前n項(xiàng)和,則下列結(jié)論正確的是(

)A.?dāng)?shù)列SKIPIF1<0的第SKIPIF1<0項(xiàng)為SKIPIF1<0 B.?dāng)?shù)列SKIPIF1<0的第2023項(xiàng)為SKIPIF1<0C.?dāng)?shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0 D.SKIPIF1<0【答案】ACD【分析】由數(shù)列的定義,對(duì)通項(xiàng)和前n項(xiàng)和的性質(zhì)進(jìn)行討論,驗(yàn)證選項(xiàng)是否正確.【詳解】SKIPIF1<0SKIPIF1<0…,SKIPIF1<0,故A選項(xiàng)正確;SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故B選項(xiàng)錯(cuò)誤;SKIPIF1<0,SKIPIF1<0,…,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,故C選項(xiàng)正確;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,故D選項(xiàng)正確;故選:ACD.【點(diǎn)睛】方法點(diǎn)睛:解決新定義問(wèn)題,首先考查對(duì)定義的理解。其次是考查滿足新定義的數(shù)列的簡(jiǎn)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(shí)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(xiàng)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(duì)A:取特值,結(jié)合SKIPIF1<0,運(yùn)算求解即可;對(duì)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(duì)C:可得SKIPIF1<0,作差即可得結(jié)果;對(duì)D:利用累加法求得SKIPIF1<0,整理可得SKIPIF1<0,結(jié)合對(duì)勾函數(shù)的單調(diào)性分析運(yùn)算.【詳解】由SKIPIF1<0,可得:對(duì)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(cuò)誤;對(duì)B:對(duì)SKIPIF1<0,則SKIPIF1<0,故數(shù)列SKIPIF1<0為遞增數(shù)列,B正確;對(duì)C:當(dāng)SKIPIF1<0時(shí),可得SKIPIF1<0,則SKIPIF1<0,故數(shù)列SKIPIF1<0為等差數(shù)列,C正確;對(duì)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(duì)SKIPIF1<0恒成立,故當(dāng)SKIPIF1<0時(shí),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(wèn)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(gè)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(shí),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(wèn)題.(3)利用函數(shù)的方法研究數(shù)列中相關(guān)問(wèn)題時(shí),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(cè))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(lái)研究數(shù),他們根據(jù)沙粒或小石子所排列的形狀,把數(shù)分成許多類,如圖中第一行圖形中黑色小點(diǎn)個(gè)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(gè)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(shuō)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(xiàng),再逐一分析各個(gè)選項(xiàng)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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