新高考數(shù)學(xué)二輪復(fù)習(xí)重難點(diǎn)5-2 數(shù)列前n項(xiàng)和的求法(8題型+滿分技巧+限時(shí)檢測(cè))(原卷版)_第1頁
新高考數(shù)學(xué)二輪復(fù)習(xí)重難點(diǎn)5-2 數(shù)列前n項(xiàng)和的求法(8題型+滿分技巧+限時(shí)檢測(cè))(原卷版)_第2頁
新高考數(shù)學(xué)二輪復(fù)習(xí)重難點(diǎn)5-2 數(shù)列前n項(xiàng)和的求法(8題型+滿分技巧+限時(shí)檢測(cè))(原卷版)_第3頁
新高考數(shù)學(xué)二輪復(fù)習(xí)重難點(diǎn)5-2 數(shù)列前n項(xiàng)和的求法(8題型+滿分技巧+限時(shí)檢測(cè))(原卷版)_第4頁
新高考數(shù)學(xué)二輪復(fù)習(xí)重難點(diǎn)5-2 數(shù)列前n項(xiàng)和的求法(8題型+滿分技巧+限時(shí)檢測(cè))(原卷版)_第5頁
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重難點(diǎn)5-2數(shù)列前n項(xiàng)和的求法數(shù)列求和是高考數(shù)學(xué)的必考內(nèi)容,一般利用等差數(shù)列的通項(xiàng)來構(gòu)建考查裂項(xiàng)求和,構(gòu)建等差等比數(shù)列考查錯(cuò)位相減法求和,解答題中等差數(shù)列、等比數(shù)列通項(xiàng)的考查往往是第1問,數(shù)列求和則是第2問。近幾年在數(shù)列求和中加大了思維能力的考查,減少了對(duì)程序化計(jì)算(錯(cuò)位相減、裂項(xiàng)相消)的考查,主要基于新的情景,要求考生通過歸納或挖掘數(shù)列各項(xiàng)間關(guān)系發(fā)現(xiàn)規(guī)律再進(jìn)行求和。【題型1公式法求數(shù)列前n項(xiàng)和】滿分技巧(1)等差數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0,推導(dǎo)方法:倒序相加法.(2)等比數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0,推導(dǎo)方法:乘公比,錯(cuò)位相減法.(3)一些常見的數(shù)列的前n項(xiàng)和:①SKIPIF1<0;SKIPIF1<0②SKIPIF1<0;③SKIPIF1<0;=4\*GB3④SKIPIF1<0【例1】(2023·廣東珠?!そy(tǒng)考模擬預(yù)測(cè))已知SKIPIF1<0為等比數(shù)列,且SKIPIF1<0,若SKIPIF1<0.(1)求數(shù)列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.【變式1-1】(2023·寧夏銀川·高三校聯(lián)考階段練習(xí))設(shè)正項(xiàng)等比數(shù)列SKIPIF1<0且SKIPIF1<0的等差中項(xiàng)為SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)若SKIPIF1<0,數(shù)列SKIPIF1<0的前n項(xiàng)為SKIPIF1<0,數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,求SKIPIF1<0.【變式1-2】(2023·山西·??寄M預(yù)測(cè))已知等差數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0,若SKIPIF1<0,求SKIPIF1<0的最小值.【變式1-3】(2023·四川德陽·統(tǒng)考一模)已知首項(xiàng)為SKIPIF1<0的等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0成等差數(shù)列.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)求數(shù)列SKIPIF1<0的最大項(xiàng).【變式1-4】(2023·山西臨汾·??寄M預(yù)測(cè))在數(shù)列SKIPIF1<0中,SKIPIF1<0,且SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0為SKIPIF1<0的前n項(xiàng)和,求使得SKIPIF1<0成立的最小正整數(shù)n的值.【題型2分組法求數(shù)列前n項(xiàng)和】滿分技巧(1)適用范圍:某些數(shù)列的求和是將數(shù)列轉(zhuǎn)化為若干個(gè)可求和的新數(shù)列的和或差,從而求得原數(shù)列的和,注意在含有字母的數(shù)列中對(duì)字母的討論.(2)常見類型:=1\*GB3①若an=bn±cn,且{bn},{cn}為等差或等比數(shù)列;=2\*GB3②通項(xiàng)公式為an=eq\b\lc\{(\a\vs4\al\co1(bn,n為奇數(shù),,cn,n為偶數(shù)))的數(shù)列,其中數(shù)列{bn},{cn}是等比數(shù)列或等差數(shù)列.【例2】(2023·山西忻州·高三校聯(lián)考階段練習(xí))已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0(SKIPIF1<0).(1)求SKIPIF1<0的通項(xiàng)公式;(2)設(shè)數(shù)列SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0.【變式2-1】(2023·江蘇無錫·高三校聯(lián)考階段練習(xí))已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【變式2-2】(2023·江西貴溪·高三貴溪市實(shí)驗(yàn)中學(xué)校聯(lián)考階段練習(xí))已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,SKIPIF1<0,等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)令SKIPIF1<0,求數(shù)列SKIPIF1<0的前10項(xiàng)和.【變式2-3】(2023·廣東廣州·統(tǒng)考模擬預(yù)測(cè))設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)若數(shù)列SKIPIF1<0滿足SKIPIF1<0,求數(shù)列SKIPIF1<0的前2n項(xiàng)和SKIPIF1<0.【變式2-4】(2023·山東濰坊·統(tǒng)考模擬預(yù)測(cè))已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且滿足SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)數(shù)列SKIPIF1<0滿足SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【題型3并項(xiàng)法求數(shù)列前n項(xiàng)和】滿分技巧一個(gè)數(shù)列的前n項(xiàng)和中,可兩兩結(jié)合求解,則稱之為并項(xiàng)求和.形如an=(-1)nf(n)類型,可采用兩項(xiàng)合并求解.例如,SKIPIF1<0.【例3】(2023·陜西西安·高三??茧A段練習(xí))若數(shù)列SKIPIF1<0的通項(xiàng)公式是SKIPIF1<0,則該數(shù)列的前100項(xiàng)之和為.【變式3-1】(2023·河北邯鄲·統(tǒng)考模擬預(yù)測(cè))已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且滿足SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)若數(shù)列SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【變式3-2】(2023·廣東廣州·高三統(tǒng)考階段練習(xí))記SKIPIF1<0為等差數(shù)列SKIPIF1<0的前n項(xiàng)和,已知SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)記SKIPIF1<0,求數(shù)列SKIPIF1<0的前23項(xiàng)的和SKIPIF1<0.【變式3-3】(2023·湖南邵陽·高三校聯(lián)考階段練習(xí))已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列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.【變式3-4】(2023·重慶·高三重慶一中??茧A段練習(xí))已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0.(1)求證:數(shù)列SKIPIF1<0為等比數(shù)列;(2)若SKIPIF1<0,求數(shù)列SKIPIF1<0的前n項(xiàng)的和SKIPIF1<0.【題型4逆序相加法求數(shù)列前n項(xiàng)和】滿分技巧如果一個(gè)數(shù)列{an}的前n項(xiàng)中首末兩端等“距離”的兩項(xiàng)的和相等或等于同一個(gè)常數(shù),那么求這個(gè)數(shù)列的前n項(xiàng)和即可用倒序相加法,如等差數(shù)列的前n項(xiàng)和公式即是用此法推導(dǎo)的.【例4】(2023·重慶·高三重慶一中??茧A段練習(xí))已知SKIPIF1<0為正項(xiàng)等比數(shù)列,且SKIPIF1<0,若函數(shù)SKIPIF1<0,則SKIPIF1<0()A.2023B.2024C.SKIPIF1<0D.1012【變式4-1】(2023·山東濰坊·高三安丘市第一中學(xué)校考階段練習(xí))已知函數(shù)SKIPIF1<0,數(shù)列SKIPIF1<0為等比數(shù)列,SKIPIF1<0,且SKIPIF1<0,利用課本中推導(dǎo)等差數(shù)列前SKIPIF1<0項(xiàng)和的公式的方法,則SKIPIF1<0()A.SKIPIF1<0B.2017C.4034D.8068【變式4-2】(2023·全國·本溪高中校聯(lián)考模擬預(yù)測(cè))“數(shù)學(xué)王子”高斯是近代數(shù)學(xué)奠基者之一,他的數(shù)學(xué)研究幾乎遍及所有領(lǐng)域,在數(shù)論?代數(shù)學(xué)?非歐幾何?復(fù)變函數(shù)和微分幾何等方面都作出了開創(chuàng)性的貢獻(xiàn).我們高中階段也學(xué)習(xí)過很多高斯的數(shù)學(xué)理論,比如高斯函數(shù)?倒序相加法?最小二乘法等等.已知某數(shù)列的通項(xiàng)SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【變式4-3】(2023·全國·高三專題練習(xí))已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0,設(shè)函數(shù)SKIPIF1<0,則SKIPIF1<0.【變式4-4】(2023·云南·高三云南師大附中校考階段練習(xí))已知數(shù)列SKIPIF1<0滿足:SKIPIF1<0(SKIPIF1<0),數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)求SKIPIF1<0.【題型5錯(cuò)位相減法求數(shù)列前n項(xiàng)和】滿分技巧1、解題步驟2、注意解題“3關(guān)鍵”①要善于識(shí)別題目類型,特別是等比數(shù)列公比為負(fù)數(shù)的情形.②在寫出“Sn”與“qSn”的表達(dá)式時(shí)應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對(duì)齊”以便下一步準(zhǔn)確寫出“Sn-qSn”的表達(dá)式.③在應(yīng)用錯(cuò)位相減法求和時(shí),若等比數(shù)列的公比為參數(shù),應(yīng)分公比q=1和q≠1兩種情況求解.3、等差乘等比數(shù)列求和,令SKIPIF1<0,可以用錯(cuò)位相減法.SKIPIF1<0①SKIPIF1<0②SKIPIF1<0得:SKIPIF1<0.整理得:SKIPIF1<0.【例5】(2023·江蘇鹽城·高三鹽城中學(xué)校聯(lián)考階段練習(xí))已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,且數(shù)列SKIPIF1<0是等差數(shù)列.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【變式5-1】(2023·青?!ばB?lián)考模擬預(yù)測(cè))已知數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)求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.【變式5-2】(2023·山東泰安·高三統(tǒng)考期中)已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,SKIPIF1<0且SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0;(2)記SKIPIF1<0,求數(shù)列SKIPIF1<0的前n項(xiàng)和.【變式5-3】(2023·海南·校聯(lián)考模擬預(yù)測(cè))已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0.(1)求SKIPIF1<0;(2)若SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【變式5-4】(2023·江蘇南京·高三期末)已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,且對(duì)任意SKIPIF1<0都有SKIPIF1<0SKIPIF1<0.(1)設(shè)SKIPIF1<0,證明:SKIPIF1<0是等差數(shù)列;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【題型6裂項(xiàng)相消法求數(shù)列前n項(xiàng)和】滿分技巧1、用裂項(xiàng)法求和的裂項(xiàng)原則及規(guī)律(1)裂項(xiàng)原則:一般是前邊裂幾項(xiàng),后邊就裂幾項(xiàng),直到發(fā)現(xiàn)被消去項(xiàng)的規(guī)律為止.(2)消項(xiàng)規(guī)律:消項(xiàng)后前邊剩幾項(xiàng),后邊就剩幾項(xiàng),前邊剩第幾項(xiàng),后邊就剩倒數(shù)第幾項(xiàng).【注意】利用裂項(xiàng)相消法求和時(shí),既要注意檢驗(yàn)通項(xiàng)公式裂項(xiàng)前后是否等價(jià),又要注意求和時(shí),正負(fù)項(xiàng)相消消去了哪些項(xiàng),保留了哪些項(xiàng),切不可漏寫未被消去的項(xiàng).2、裂項(xiàng)相消法中常見的裂項(xiàng)技巧(1)(2)(3)(4)(5)(6)(7)【例6】(2023·四川南充·統(tǒng)考一模)已知數(shù)列SKIPIF1<0是首項(xiàng)為2的等比數(shù)列,公比SKIPIF1<0,且SKIPIF1<0是SKIPIF1<0和SKIPIF1<0的等差中項(xiàng).(1)求SKIPIF1<0的通項(xiàng)公式;(2)設(shè)數(shù)列SKIPIF1<0滿足SKIPIF1<0,求SKIPIF1<0的前2023項(xiàng)和SKIPIF1<0.【變式6-1】(2023·江蘇鎮(zhèn)江·高三校考階段練習(xí))已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,SKIPIF1<0是n、SKIPIF1<0的等差中項(xiàng),SKIPIF1<0.(1)證明:SKIPIF1<0是等比數(shù)列;(2)設(shè)SKIPIF1<0,數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0,證明:SKIPIF1<0.【變式6-2】(2023·福建莆田·高三莆田第四中學(xué)校考階段練習(xí))已知數(shù)列SKIPIF1<0前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且滿足SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0,求證:SKIPIF1<0.【變式6-3】(2023·廣東珠?!じ呷楹J械谝恢袑W(xué)校考期末)已知正項(xiàng)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,SKIPIF1<0,且當(dāng)SKIPIF1<0時(shí)SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)若數(shù)列SKIPIF1<0滿足SKIPIF1<0,數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,試比較SKIPIF1<0與SKIPIF1<0的大小,并加以證明.【變式6-4】(2023·河北保定·高三校聯(lián)考階段練習(xí))設(shè)SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)SKIPIF1<0,證明:SKIPIF1<0.【題型7含絕對(duì)值數(shù)列的前n項(xiàng)和】【例7】(2023·湖北武漢·統(tǒng)考模擬預(yù)測(cè))已知SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列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.【變式7-1】(2023·遼寧丹東·高三校聯(lián)考階段練習(xí))已知等差數(shù)列SKIPIF1<0的公差為整數(shù),SKIPIF1<0,設(shè)其前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0是公差為SKIPIF1<0的等差數(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的前n項(xiàng)和SKIPIF1<0.【變式7-2】(2023·重慶·高三重慶市第七中學(xué)校校考階段練習(xí))已知SKIPIF1<0是正項(xiàng)等比數(shù)列.SKIPIF1<0,且SKIPIF1<0,(1)求SKIPIF1<0的通項(xiàng)公式;(2)當(dāng)SKIPIF1<0為遞增數(shù)列,設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.【變式7-3】(2023·陜西西安·高三統(tǒng)考階段練習(xí))已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,且SKIPIF1<0.(1)求SKIPIF1<0的通項(xiàng)公式;(2)記SKIPIF1<0,求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0.【變式7-4】(2023·全國·模擬預(yù)測(cè))在數(shù)列SKIPIF1<0中,SKIPIF1<0,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.【題型8數(shù)列求和與不等式綜合】滿分技巧常見的角度主要包括兩個(gè)方面:一、不等式恒成立小件下,求參數(shù)的取值范圍;二、不等式的證明,常見方法有不比較法、構(gòu)造輔助函數(shù)法、放縮法、數(shù)學(xué)歸納法等?!纠?】(2023·河南·信陽高中校聯(lián)考模擬預(yù)測(cè))已知SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,且SKIPIF1<0為正項(xiàng)等比數(shù)列,SKIPIF1<0,SKIPIF1<0.(1)求證:數(shù)列SKIPIF1<0是等差數(shù)列;(2)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(3)設(shè)SKIPIF1<0,且數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,若SKIPIF1<0恒成立,求實(shí)數(shù)SKIPIF1<0的取值范圍.【變式8-1】(2023·山東·山東省五蓮縣第一中學(xué)校聯(lián)考模擬預(yù)測(cè))已知數(shù)列SKIPIF1<0前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且對(duì)任意的正整數(shù)SKIPIF1<0與SKIPIF1<0的等差中項(xiàng)為SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)證明:SKIPIF1<0.【變式8-2】(2023·安徽·高三校聯(lián)考階段練習(xí))已知數(shù)列SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0,數(shù)列SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0(SKIPIF1<0表示不超過SKIPIF1<0的最達(dá)整數(shù)),SKIPIF1<0.(1)求SKIPIF1<0;(2)令SKIPIF1<0,記數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,求證:SKIPIF1<0.【變式8-3】(2023·河北石家莊·高三校聯(lián)考期末)已知數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)若SKIPIF1<0為等差數(shù)列,求SKIPIF1<0的通項(xiàng)公式;(2)記SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立,求SKIPIF1<0的取值范圍.【變式8-4】(2023·山東青島·高二山東省青島第五十八中學(xué)??计谀┮阎瘮?shù)SKIPIF1<0滿足SKIPIF1<0,若數(shù)列SKIPIF1<0滿足:SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)若數(shù)列SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0(SKIPIF1<0),數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0,若SKIPIF1<0對(duì)一切SKIPIF1<0恒成立,求實(shí)數(shù)SKIPIF1<0的取值范圍.(建議用時(shí):60分鐘)1.(2023·全國·高三專題練習(xí))已知數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0,SKIPIF1<0為數(shù)列的前n項(xiàng)和,SKIPIF1<0()A.1009B.1010C.1011D.10122.(2023·湖南長(zhǎng)沙·高三周南中學(xué)??奸_學(xué)考試)已知函數(shù)SKIPIF1<0,在正項(xiàng)等比數(shù)列SKIPIF1<0中,SKIPIF1<0,則SKIPIF1<0()A.1011B.1012C.2023D.20243.(2023·天津·高三南開中學(xué)??茧A段練習(xí))在公差大于0的等差數(shù)列SKIPIF1<0中,SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,則數(shù)列SKIPIF1<0的前21項(xiàng)和為(

)A.12B.21C.11D.314.(2023·天津·高三統(tǒng)考期中)設(shè)等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,數(shù)列SKIPIF1<0的前SKIPIF1<0和為SKIPIF1<0,已知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則正整數(shù)SKIPIF1<0的值為()A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<05.(2023·廣西·模擬預(yù)測(cè))設(shè)SKIPIF1<0是等差數(shù)列,SKIPIF1<0是各項(xiàng)都為正數(shù)的等比數(shù)列.且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.(1)求SKIPIF1<0,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<0.6.(2023·貴州貴陽·高三貴陽一中校考階段練習(xí))數(shù)列SKIPIF1<0滿足SKIPIF1<0,

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