人教A版高中數(shù)學(xué)(選擇性必修二)同步講義第08講 第四章 數(shù)列 重點(diǎn)題型章末總結(jié)(教師版)_第1頁(yè)
人教A版高中數(shù)學(xué)(選擇性必修二)同步講義第08講 第四章 數(shù)列 重點(diǎn)題型章末總結(jié)(教師版)_第2頁(yè)
人教A版高中數(shù)學(xué)(選擇性必修二)同步講義第08講 第四章 數(shù)列 重點(diǎn)題型章末總結(jié)(教師版)_第3頁(yè)
人教A版高中數(shù)學(xué)(選擇性必修二)同步講義第08講 第四章 數(shù)列 重點(diǎn)題型章末總結(jié)(教師版)_第4頁(yè)
人教A版高中數(shù)學(xué)(選擇性必修二)同步講義第08講 第四章 數(shù)列 重點(diǎn)題型章末總結(jié)(教師版)_第5頁(yè)
已閱讀5頁(yè),還剩22頁(yè)未讀 繼續(xù)免費(fèi)閱讀

下載本文檔

版權(quán)說(shuō)明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)

文檔簡(jiǎn)介

第08講第四章數(shù)列重點(diǎn)題型章末總結(jié)一、思維導(dǎo)圖二、題型精講題型01等差與等比數(shù)列的基本運(yùn)算1.(2023·全國(guó)·高二隨堂練習(xí))已知數(shù)列SKIPIF1<0為等差數(shù)列,前n項(xiàng)和為SKIPIF1<0,求解下列問(wèn)題:(1)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(2)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(3)若SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,求n.【答案】(1)2(2)1596(3)11【詳解】(1)由題意知數(shù)列SKIPIF1<0為等差數(shù)列,SKIPIF1<0,SKIPIF1<0,設(shè)SKIPIF1<0公差為d,故SKIPIF1<0,解得SKIPIF1<0;(2)數(shù)列SKIPIF1<0為等差數(shù)列,SKIPIF1<0,SKIPIF1<0,設(shè)SKIPIF1<0公差為d,故SKIPIF1<0,解得SKIPIF1<0,則SKIPIF1<0;(3)由題意知數(shù)列SKIPIF1<0為等差數(shù)列,SKIPIF1<0,SKIPIF1<0,設(shè)SKIPIF1<0公差為d,則SKIPIF1<0,解得SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0(舍去),故SKIPIF1<0.2.(2023秋·高二課時(shí)練習(xí))在等差數(shù)列SKIPIF1<0中,(1)已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(2)已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(3)已知SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(4)已知SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(4)SKIPIF1<0【詳解】(1)由SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.(2)由SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0.(3)由SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.(4)由SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.3.(2023·全國(guó)·高二課堂例題)已知數(shù)列SKIPIF1<0是等差數(shù)列.(1)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(2)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(3)若SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,求n.【答案】(1)2700(2)SKIPIF1<0(3)SKIPIF1<0.【詳解】(1)因?yàn)镾KIPIF1<0,SKIPIF1<0,根據(jù)公式SKIPIF1<0,可得SKIPIF1<0.(2)因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0.根據(jù)公式SKIPIF1<0,可得SKIPIF1<0.(3)把SKIPIF1<0,SKIPIF1<0,SKIPIF1<0代入SKIPIF1<0,得SKIPIF1<0.整理,得SKIPIF1<0.解得SKIPIF1<0,或SKIPIF1<0(舍去).所以SKIPIF1<0.4.(2023秋·高二課時(shí)練習(xí))在等差數(shù)列SKIPIF1<0中,(1)已知SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(2)已知SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0.【答案】(1)44(2)SKIPIF1<0【詳解】(1)設(shè)SKIPIF1<0,則SKIPIF1<0解得SKIPIF1<0故SKIPIF1<0.(2)設(shè)SKIPIF1<0,則SKIPIF1<0解得SKIPIF1<0故SKIPIF1<0.5.(2023秋·云南·高三校聯(lián)考階段練習(xí))在正項(xiàng)等比數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)若SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】(1)設(shè)正項(xiàng)等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0,則SKIPIF1<0.由SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0.(2)由(1)可知SKIPIF1<0,則SKIPIF1<0是以1為首項(xiàng),2為公差的等差數(shù)列,故SKIPIF1<0.6.(2023·全國(guó)·高二隨堂練習(xí))已知數(shù)列SKIPIF1<0為等比數(shù)列.(1)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(2)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0和q;(3)若SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0或SKIPIF1<0【詳解】(1)因?yàn)閿?shù)列SKIPIF1<0為等比數(shù)列,且SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,(2)因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,(3)因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,由題意可知SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,綜上SKIPIF1<0或SKIPIF1<07.(2023·全國(guó)·高二隨堂練習(xí))求下列等比數(shù)列SKIPIF1<0的前n項(xiàng)和.(1)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0;(2)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0;(3)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0;(4)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0(4)378【詳解】(1)由SKIPIF1<0,SKIPIF1<0,SKIPIF1<0得SKIPIF1<0(2)由SKIPIF1<0,SKIPIF1<0,SKIPIF1<0得SKIPIF1<0(3)由SKIPIF1<0,SKIPIF1<0,SKIPIF1<0得SKIPIF1<0(4)由SKIPIF1<0,SKIPIF1<0,SKIPIF1<0得SKIPIF1<08.(2023·全國(guó)·高二隨堂練習(xí))已知數(shù)列SKIPIF1<0為等比數(shù)列,前n項(xiàng)和為SKIPIF1<0.(1)如果SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0;(2)如果SKIPIF1<0,SKIPIF1<0,求q;(3)如果SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0(3)SKIPIF1<0【詳解】(1)SKIPIF1<0等比數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,解得SKIPIF1<0.(2)在等比數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,顯然公比SKIPIF1<0,SKIPIF1<0SKIPIF1<0,整理得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.(3)因?yàn)镾KIPIF1<0,SKIPIF1<0,所以公比SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0.題型02等差、等比數(shù)列的判定1.(2023春·山東淄博·高二??茧A段練習(xí))已知下列數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0的公式SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)判斷該數(shù)列是否為等差數(shù)列,并說(shuō)明理由.【答案】(1)SKIPIF1<0(2)不是等差數(shù)列,理由見(jiàn)解析【詳解】(1)因?yàn)镾KIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),上式不成立,所以SKIPIF1<0;(2)由(1)得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以數(shù)列SKIPIF1<0不是等差數(shù)列.2.(2023春·云南曲靖·高一曲靖一中??计谀?shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0是常數(shù).(1)當(dāng)SKIPIF1<0時(shí),求SKIPIF1<0及SKIPIF1<0的值;(2)數(shù)列SKIPIF1<0是否可能為等差數(shù)列?若可能,求出它的通項(xiàng)公式;若不可能,說(shuō)明理由;【答案】(1)SKIPIF1<0,SKIPIF1<0(2)數(shù)列SKIPIF1<0不是等差數(shù)列,理由見(jiàn)解析【詳解】(1)因?yàn)镾KIPIF1<0,又SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0;(2)因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0,假設(shè)數(shù)列SKIPIF1<0是等差數(shù)列,則SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,數(shù)列SKIPIF1<0不是等差數(shù)列,故假設(shè)不成立,故數(shù)列SKIPIF1<0不可能為等差數(shù)列3.(2023春·上海嘉定·高二統(tǒng)考期末)已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)求證:數(shù)列SKIPIF1<0是等差數(shù)列.【答案】(1)SKIPIF1<0(2)見(jiàn)解析【詳解】(1)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,所以SKIPIF1<0.(2)由(1)知,SKIPIF1<0,所以數(shù)列SKIPIF1<0是以首項(xiàng)為SKIPIF1<0,公差為SKIPIF1<0等差數(shù)列.4.(2023春·貴州銅仁·高二統(tǒng)考期末)已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0SKIPIF1<0.(1)求證:數(shù)列SKIPIF1<0是等比數(shù)列;(2)求數(shù)列SKIPIF1<0的通項(xiàng)公式及它的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【答案】(1)證明見(jiàn)解析(2)SKIPIF1<0,SKIPIF1<0【詳解】(1)證明:因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以數(shù)列SKIPIF1<0是以3為首項(xiàng),3為公比的等比數(shù)列.(2)由(1)得SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<05.(2023·全國(guó)·高二專題練習(xí))已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.證明:數(shù)列SKIPIF1<0為等比數(shù)列;【答案】證明見(jiàn)解析【詳解】(1)由題意,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0.由題意知SKIPIF1<0,①

當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,②①-②得SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0.則SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0所以SKIPIF1<0是以SKIPIF1<0為首項(xiàng),2為公比的等比數(shù)列.6.(2023秋·黑龍江大慶·高三大慶市東風(fēng)中學(xué)??茧A段練習(xí))在數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0.(1)求證:數(shù)列SKIPIF1<0為等比數(shù)列,并求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【答案】(1)證明見(jiàn)解析;SKIPIF1<0;(2)SKIPIF1<0【詳解】(1)SKIPIF1<0,SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,數(shù)列SKIPIF1<0是首項(xiàng)為SKIPIF1<0,公比為SKIPIF1<0的等比數(shù)列,SKIPIF1<0,SKIPIF1<0;(2)SKIPIF1<0數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0SKIPIF1<0.7.(2023·全國(guó)·高二專題練習(xí))在數(shù)列SKIPIF1<0中,已知SKIPIF1<0,且SKIPIF1<0.(1)求證:數(shù)列SKIPIF1<0是等比數(shù)列.(2)求數(shù)列SKIPIF1<0的通項(xiàng)公式.【答案】(1)證明見(jiàn)解析(2)SKIPIF1<0【詳解】(1)令SKIPIF1<0,∴SKIPIF1<0,∵SKIPIF1<0,故SKIPIF1<0,∴數(shù)列SKIPIF1<0是公比為2的等比數(shù)列,即數(shù)列SKIPIF1<0是公比為2的等比數(shù)列.(2)由(1)易知SKIPIF1<0,即SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0.題型03等差、等比數(shù)列的性質(zhì)及應(yīng)用1.(2023秋·天津河?xùn)|·高三天津市第四十五中學(xué)??茧A段練習(xí))若數(shù)列SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0,則其前17項(xiàng)和SKIPIF1<0(

)A.136 B.119 C.102 D.85【答案】B【詳解】根據(jù)SKIPIF1<0可得SKIPIF1<0,所以數(shù)列SKIPIF1<0是公差為2的等差數(shù)列,利用等差數(shù)列性質(zhì)由SKIPIF1<0可得SKIPIF1<0;所以其前17項(xiàng)和SKIPIF1<0.故選:B2.(2023春·新疆伊犁·高二統(tǒng)考期中)記等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(

)A.6 B.7 C.8 D.9【答案】D【詳解】解:因?yàn)镾KIPIF1<0為等差數(shù)列,SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0(SKIPIF1<0為等差數(shù)列SKIPIF1<0的公差).故選:D.3.(2023秋·吉林白城·高三校考階段練習(xí))已知等差數(shù)列SKIPIF1<0是遞增數(shù)列,且滿足SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0等于(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【詳解】由等差數(shù)列的性質(zhì),SKIPIF1<0,又SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,又SKIPIF1<0是遞增數(shù)列,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.故選:C.4.(2023春·河南周口·高二校聯(lián)考期中)設(shè)等差數(shù)列SKIPIF1<0,SKIPIF1<0的前SKIPIF1<0項(xiàng)和分別為SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【詳解】因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0.故選:D5.(2023春·新疆·高二八一中學(xué)??计谀┤魞蓚€(gè)等差數(shù)列SKIPIF1<0,SKIPIF1<0的前n項(xiàng)和SKIPIF1<0滿足SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【詳解】由SKIPIF1<0,得SKIPIF1<0.故選:B.6.(2023·河北唐山·開(kāi)灤第二中學(xué)??寄M預(yù)測(cè))已知等差數(shù)列SKIPIF1<0(SKIPIF1<0)的前n項(xiàng)和為SKIPIF1<0,公差SKIPIF1<0,SKIPIF1<0,則使得SKIPIF1<0的最大整數(shù)n為(

)A.9 B.10 C.17 D.18【答案】C【詳解】解:因?yàn)镾KIPIF1<0,所以SKIPIF1<0異號(hào),因?yàn)镾KIPIF1<0,所以SKIPIF1<0,又有SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0的最大整數(shù)n為17.故選:C7.(2023秋·云南昆明·高三云南民族大學(xué)附屬中學(xué)??茧A段練習(xí))已知等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(

)A.8 B.9 C.16 D.17【答案】A【詳解】設(shè)SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0為等比數(shù)列,所以SKIPIF1<0仍成等比數(shù)列.易知SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0.故選:A.8.(2023秋·云南·高三云南師大附中校考階段練習(xí))已知等比數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(

)A.90 B.135 C.150 D.180【答案】C【詳解】由題意,在等比數(shù)列SKIPIF1<0中,SKIPIF1<0,由等比數(shù)列前n項(xiàng)和的性質(zhì)可得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,∴有SKIPIF1<0,即SKIPIF1<0,整理可得SKIPIF1<0,解得SKIPIF1<0(舍)或SKIPIF1<0,∵SKIPIF1<0,∴有SKIPIF1<0,解得SKIPIF1<0,故選:C.9.(2023·福建泉州·統(tǒng)考模擬預(yù)測(cè))記等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0.若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【詳解】設(shè)等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0(SKIPIF1<0),則SKIPIF1<0,解得:SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,故選:C.10.(2023·河南·模擬預(yù)測(cè))已知等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.【答案】SKIPIF1<0【詳解】設(shè)等差數(shù)列SKIPIF1<0的公差為SKIPIF1<0,由等差數(shù)列前n項(xiàng)和公式可知SKIPIF1<0;可得SKIPIF1<0為定值,所以SKIPIF1<0即為等差數(shù)列,又SKIPIF1<0,即SKIPIF1<0是以SKIPIF1<0為首項(xiàng),公差為1的等差數(shù)列,所以SKIPIF1<0,從而SKIPIF1<0.故答案為:SKIPIF1<011.(2023秋·福建寧德·高二福建省寧德第一中學(xué)校考開(kāi)學(xué)考試)已知等差數(shù)列SKIPIF1<0,SKIPIF1<0,其前SKIPIF1<0項(xiàng)和分別為SKIPIF1<0,SKIPIF1<0,且滿足SKIPIF1<0,SKIPIF1<0.【答案】SKIPIF1<0【詳解】運(yùn)用等差數(shù)列的性質(zhì)SKIPIF1<0,可得SKIPIF1<0即SKIPIF1<0,由等差數(shù)列性質(zhì)可知SKIPIF1<0.故答案為:SKIPIF1<0.12.(2023秋·江西南昌·高三江西師大附中校考階段練習(xí))已知數(shù)列SKIPIF1<0為等比數(shù)列,且SKIPIF1<0,則SKIPIF1<0.【答案】SKIPIF1<0【詳解】由SKIPIF1<0為等比數(shù)列,則SKIPIF1<0,又SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0.故答案為:SKIPIF1<0.13.(2023·全國(guó)·高二隨堂練習(xí))已知等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,求SKIPIF1<0.【答案】81【詳解】由于等差數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,故SKIPIF1<0也成等差數(shù)列,即SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0.14.(2023·全國(guó)·高二隨堂練習(xí))在由正數(shù)組成的等比數(shù)列SKIPIF1<0中,若SKIPIF1<0,求SKIPIF1<0的值.【答案】SKIPIF1<0【詳解】SKIPIF1<0數(shù)列SKIPIF1<0是由正數(shù)組成的等比數(shù)列,SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0,故SKIPIF1<0,則SKIPIF1<0SKIPIF1<0.題型04數(shù)列求通項(xiàng)、求和1.(2023·浙江·模擬預(yù)測(cè))已知數(shù)列SKIPIF1<0滿足SKIPIF1<0(1)若SKIPIF1<0,求數(shù)列SKIPIF1<0的通項(xiàng)SKIPIF1<0;(2)記SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)之和,若SKIPIF1<0,求SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0.【詳解】(1)當(dāng)SKIPIF1<0,SKIPIF1<0①,SKIPIF1<0②,①SKIPIF1<0②可得SKIPIF1<0,左右同時(shí)乘以SKIPIF1<0可以得出:SKIPIF1<0,即得SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0應(yīng)用累加法可得:SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0(2)由(1)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0或SKIPIF1<0.2.(2023·遼寧撫順·??寄M預(yù)測(cè))已知各項(xiàng)均為正數(shù)的數(shù)列SKIPIF1<0,SKIPIF1<0滿足:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列SKIPIF1<0,SKIPIF1<0的通項(xiàng)公式;(2)求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0.【答案】(1)SKIPIF1<0,SKIPIF1<0;(2)SKIPIF1<0【詳解】(1)由SKIPIF1<0,得SKIPIF1<0,又SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,符合上式,SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0.(2)由(1)知SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,兩式相減得SKIPIF1<0,所以SKIPIF1<0.3.(2023·湖南永州·統(tǒng)考一模)已知數(shù)列SKIPIF1<0是公比SKIPIF1<0的等比數(shù)列,前三項(xiàng)和為39,且SKIPIF1<0成等差數(shù)列.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0,求SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】(1)由題意可得SKIPIF1<0,即得SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,可得SKIPIF1<0,由于SKIPIF1<0,故得SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0;(2)由(1)結(jié)論可得SKIPIF1<0SKIPIF1<0,故SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0SKIPIF1<0.4.(2023·遼寧撫順·??寄M預(yù)測(cè))在數(shù)列SKIPIF1<0中,已知SKIPIF1<0,SKIPIF1<0,記SKIPIF1<0.(1)證明:數(shù)列SKIPIF1<0為等比數(shù)列;(2)記______,數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,求SKIPIF1<0.在①SKIPIF1<0;②SKIPIF1<0;③SKIPIF1<0三個(gè)條件中選擇一個(gè)補(bǔ)充在第(2)問(wèn)中并對(duì)其求解.注:如果選擇多個(gè)條件分別解答,按第一個(gè)解答計(jì)分.【答案】(1)證明見(jiàn)解析;(2)答案見(jiàn)解析.【詳解】(1)由SKIPIF1<0,得SKIPIF1<0,則SKIPIF1<0,而SKIPIF1<0,因此SKIPIF1<0,顯然SKIPIF1<0,所以數(shù)列SKIPIF1<0為以2為首項(xiàng),2為公比的等比數(shù)列.(2)選擇①:由(1)得,SKIPIF1<0,則SKIPIF1<0所以SKIPIF1<0.選擇②:由(1)得,SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0.選擇③:由(1)得,SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0.5.(2023·陜西商洛·陜西省丹鳳中學(xué)??寄M預(yù)測(cè))記等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,已知SKIPIF1<0,且SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】(1)設(shè)數(shù)列SKIPIF1<0的公差為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0.因?yàn)镾KIPIF1<0,所以SKIPIF1<0,整理得SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,符合題意;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,不符合題意,舍去,所以SKIPIF1<0.(2)由(1)知SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,則SKIPIF1<0,兩式相減,得SKIPIF1<0令SKIPIF1<0,則SKIPIF1<0,兩式相減,得SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.三、數(shù)學(xué)思想與方法函數(shù)方程1.(2023·上海浦東新·統(tǒng)考三模)已知數(shù)列SKIPIF1<0(SKIPIF1<0是正整數(shù))的遞推公式為SKIPIF1<0若存在正整數(shù)SKIPIF1<0,使得SKIPIF1<0,則SKIPIF1<0的最大值是.【答案】SKIPIF1<0【詳解】由題意,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0是SKIPIF1<0,公比為3的等比數(shù)列,SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0,SKIPIF1<0也成立,SKIPIF1<0;對(duì)于SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0SKIPIF1<0,考察SKIPIF1<0=SKIPIF1<0,其中SKIPIF1<0是對(duì)稱軸為SKIPIF1<0,開(kāi)口向下的拋物線,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0

,當(dāng)SKIPIF1<0時(shí)SKIPIF1<0最大,SKIPIF1<0;故答案為:SKIPIF1<0.2.(2023·山東淄博·山東省淄博實(shí)驗(yàn)中學(xué)??既#┮阎炔顢?shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0成等比數(shù)列,且公差SKIPIF1<0,數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0.(1)求SKIPIF1<0;(2)若數(shù)列SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0,設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,若對(duì)任意的SKIPIF1<0,都有SKIPIF1<0,求SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】(1)因?yàn)閿?shù)列SKIPIF1<0為等差數(shù)列,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,所以SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.(2)因?yàn)镾KIPIF1<0,所以SKIPIF1<0,兩式相減得SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.因?yàn)閷?duì)任意的SKIPIF1<0,都有SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.令SKIPIF1<0,則SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0遞增,而SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.3.(2023·山東濰坊·統(tǒng)考二模)已知等差數(shù)列SKIPIF1<0的公差為2,前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0成等比數(shù)列.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)若SKIPIF1<0,求數(shù)列SKIPIF1<0的最大項(xiàng).【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】(1)由題意知SKIPIF1<0,又因?yàn)镾KIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0.(2)由(1)知SKIPIF1<0,設(shè)SKIPIF1<0,所以SKIPIF1<0,又因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0因?yàn)楹瘮?shù)在SKIPIF1<0時(shí)遞減,所以SKIPIF1<0的最大值可能出現(xiàn)在SKIPIF1<0或SKIPIF1<0時(shí),SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0,所以數(shù)列SKIPIF1<0的最大項(xiàng)為SKIPIF1<0.4.(2023·浙江寧波·統(tǒng)考二模)已知等比數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0滿足SKIPIF1<0.(1)求首項(xiàng)SKIPIF1<0的值及SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0,求滿足SKIPIF1<0的最大正整數(shù)n的值.【答案】(1)SKIPIF1<0,SKIPIF1<0(2)11【詳解】(1)解法1:當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,因?yàn)閿?shù)列SKIPIF1<0是等比數(shù)列,所以公比為2,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,且SKIPIF1<0滿足題意,所以SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0.解法2:由題知,SKIPIF1<0,即SKIPIF1<0,由①SKIPIF1<0代入②,得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.(舍去),所以SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0.(2)由(1)得SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,即SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0時(shí)單調(diào)遞增,且SKIPIF1<0,而SKIPIF1<0,所以滿足條件的最大正整數(shù)SKIPIF1<0.分類討論思想1.(2023·河北滄州·??既#┰O(shè)公比為正數(shù)的等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,滿足SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0為數(shù)列SKIPIF1<0在區(qū)間SKIPIF1<0中的項(xiàng)的個(gè)數(shù),求數(shù)列SKIPIF1<0前100項(xiàng)的和.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】(1)設(shè)公比為SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,即SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0(舍去),得SKIPIF1<0,又SKIPIF1<0,所以數(shù)列SKIPIF1<0是首項(xiàng)為2,公比為2的等比數(shù)列,故數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0.(2)由SKIPIF1<0為數(shù)列SKIPIF1<0在區(qū)間SKIPIF1<0中的項(xiàng)的個(gè)數(shù),可知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0.∴SKIPIF1<0.∴數(shù)列SKIPIF1<0前100項(xiàng)的和為SK

溫馨提示

  • 1. 本站所有資源如無(wú)特殊說(shuō)明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
  • 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
  • 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒(méi)有圖紙預(yù)覽就沒(méi)有圖紙。
  • 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
  • 5. 人人文庫(kù)網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
  • 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
  • 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。

最新文檔

評(píng)論

0/150

提交評(píng)論