新高考數(shù)學二輪復習專題三 數(shù)列第2講 數(shù)列解答題（數(shù)列求通項） 原卷版

第2講數(shù)列解答題（數(shù)列求通項）目錄第一部分：知識強化第二部分：重難點題型突破突破一：SKIPIF1<0法突破二：SKIPIF1<0法突破三：累加法突破四：累乘法突破五：構(gòu)造法突破六：倒數(shù)法突破七：隔項等差突破八：隔項等比第三部分：沖刺重難點特訓第一部分：知識強化1、對于數(shù)列SKIPIF1<0，前SKIPIF1<0項和記為SKIPIF1<0；①SKIPIF1<0；②SKIPIF1<0②：SKIPIF1<0SKIPIF1<0法歸類角度1：已知SKIPIF1<0與SKIPIF1<0的關系；或SKIPIF1<0與SKIPIF1<0的關系用SKIPIF1<0，得到SKIPIF1<0例子：已知SKIPIF1<0，求SKIPIF1<0角度2：已知SKIPIF1<0與SKIPIF1<0的關系；或SKIPIF1<0與SKIPIF1<0的關系SKIPIF1<0替換題目中的SKIPIF1<0例子：已知SKIPIF1<0；已知SKIPIF1<0角度3：已知等式中左側(cè)含有：SKIPIF1<0作差法（類似SKIPIF1<0）例子：已知SKIPIF1<0求SKIPIF1<02、對于數(shù)列SKIPIF1<0，前SKIPIF1<0項積記為SKIPIF1<0；①SKIPIF1<0；②SKIPIF1<0①SKIPIF1<0②：SKIPIF1<0SKIPIF1<0法歸類角度1：已知SKIPIF1<0和SKIPIF1<0的關系角度1：用SKIPIF1<0，得到SKIPIF1<0例子：SKIPIF1<0的前SKIPIF1<0項之積SKIPIF1<0.角度2：已知SKIPIF1<0和SKIPIF1<0的關系角度1：用SKIPIF1<0替換題目中SKIPIF1<0例子：已知數(shù)列SKIPIF1<0的前n項積為SKIPIF1<0，且SKIPIF1<0.3、累加法（疊加法）若數(shù)列SKIPIF1<0滿足SKIPIF1<0，則稱數(shù)列SKIPIF1<0為“變差數(shù)列”，求變差數(shù)列SKIPIF1<0的通項時，利用恒等式SKIPIF1<0求通項公式的方法稱為累加法。具體步驟：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0將上述SKIPIF1<0個式子相加（左邊加左邊，右邊加右邊）得：SKIPIF1<0=SKIPIF1<0整理得：SKIPIF1<0=SKIPIF1<04、累乘法（疊乘法）若數(shù)列SKIPIF1<0滿足SKIPIF1<0，則稱數(shù)列SKIPIF1<0為“變比數(shù)列”，求變比數(shù)列SKIPIF1<0的通項時，利用SKIPIF1<0求通項公式的方法稱為累乘法。具體步驟：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0將上述SKIPIF1<0個式子相乘（左邊乘左邊，右邊乘右邊）得：SKIPIF1<0整理得：SKIPIF1<05、構(gòu)造法類型1：用“待定系數(shù)法”構(gòu)造等比數(shù)列形如SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0）的數(shù)列，可用“待定系數(shù)法”將原等式變形為SKIPIF1<0（其中：SKIPIF1<0），由此構(gòu)造出新的等比數(shù)列SKIPIF1<0，先求出SKIPIF1<0的通項，從而求出數(shù)列SKIPIF1<0的通項公式.標準模型：SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0）或SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0）類型2：用“同除法”構(gòu)造等差數(shù)列（1）形如SKIPIF1<0，可通過兩邊同除SKIPIF1<0，將它轉(zhuǎn)化為SKIPIF1<0，從而構(gòu)造數(shù)列SKIPIF1<0為等差數(shù)列，先求出SKIPIF1<0的通項，便可求得SKIPIF1<0的通項公式.（2）形如SKIPIF1<0，可通過兩邊同除SKIPIF1<0，將它轉(zhuǎn)化為SKIPIF1<0，換元令：SKIPIF1<0，則原式化為：SKIPIF1<0，先利用構(gòu)造法類型1求出SKIPIF1<0，再求出SKIPIF1<0的通項公式.（3）形如SKIPIF1<0的數(shù)列，可通過兩邊同除以SKIPIF1<0，變形為SKIPIF1<0的形式，從而構(gòu)造出新的等差數(shù)列SKIPIF1<0，先求出SKIPIF1<0的通項，便可求得SKIPIF1<0的通項公式.6、倒數(shù)法用“倒數(shù)變換法”構(gòu)造等差數(shù)列類型1：形如SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0）的數(shù)列，通過兩邊取“倒”，變形為SKIPIF1<0，即：SKIPIF1<0，從而構(gòu)造出新的等差數(shù)列SKIPIF1<0，先求出SKIPIF1<0的通項，即可求得SKIPIF1<0.類型2：形如SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0）的數(shù)列，通過兩邊取“倒”，變形為SKIPIF1<0，可通過換元：SKIPIF1<0，化簡為：SKIPIF1<0（此類型符構(gòu)造法類型1：用“待定系數(shù)法”構(gòu)造等比數(shù)列：形如SKIPIF1<0（SKIPIF1<0為常數(shù)，SKIPIF1<0）的數(shù)列，可用“待定系數(shù)法”將原等式變形為SKIPIF1<0（其中：SKIPIF1<0），由此構(gòu)造出新的等比數(shù)列SKIPIF1<0，先求出SKIPIF1<0的通項，從而求出數(shù)列SKIPIF1<0的通項公式.）7、隔項等差數(shù)列已知數(shù)列SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0SKIPIF1<0（其中SKIPIF1<0為常數(shù)）；或SKIPIF1<0則稱數(shù)列SKIPIF1<0為隔項等差數(shù)列，其中：①SKIPIF1<0構(gòu)成以SKIPIF1<0為首項的等差數(shù)列，公差為SKIPIF1<0；②SKIPIF1<0構(gòu)成以SKIPIF1<0為首項的等差數(shù)列，公差為SKIPIF1<0；8、隔項等比數(shù)列已知數(shù)列SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0SKIPIF1<0（其中SKIPIF1<0為常數(shù)）；或SKIPIF1<0則稱數(shù)列SKIPIF1<0為隔項等比數(shù)列，其中：①SKIPIF1<0構(gòu)成以SKIPIF1<0為首項的等比數(shù)列，公比為SKIPIF1<0；②SKIPIF1<0構(gòu)成以SKIPIF1<0為首項的等比數(shù)列，公比為SKIPIF1<0；第二部分：重難點題型突破突破一：SKIPIF1<0法1．（2022·河北張家口·高三期中）已知正項數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，其中SKIPIF1<0．(1)求SKIPIF1<0的通項公式，并判斷SKIPIF1<0是否是等差數(shù)列，說明理由；2．（2022·湖南益陽·高二階段練習）已知各項為正數(shù)的數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式；3．（2022·江蘇·蘇州中學模擬預測）已知正項數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0，且SKIPIF1<0.(1)證明：數(shù)列SKIPIF1<0為等差數(shù)列；4．（2022·江蘇南通·高三期中）已知SKIPIF1<0為正項數(shù)列SKIPIF1<0的前n項和，且SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0.(1)證明SKIPIF1<0為等差數(shù)列，并求SKIPIF1<0的通項公式；5．（2022·甘肅·高臺縣第一中學模擬預測（文））已知數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)求SKIPIF1<0的通項公式；6．（2022·全國·高三階段練習（理））已知數(shù)列SKIPIF1<0的各項均為正數(shù)，且對任意的SKIPIF1<0都有SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項公式；7．（2022·福建·泉州五中高三期中）設各項均為正數(shù)的數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0．且SKIPIF1<0，SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項公式；8．（2022·湖北襄陽·高三期中）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0(1)求SKIPIF1<0的通項公式；9．（2022·全國·高三專題練習）已知數(shù)列SKIPIF1<0的前n項和SKIPIF1<0，滿足SKIPIF1<0，SKIPIF1<0.求證：數(shù)列SKIPIF1<0是等差數(shù)列.突破二：SKIPIF1<0法1．（2022·江蘇南通·高三階段練習）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為其數(shù)列SKIPIF1<0的前SKIPIF1<0項積，且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式；2．（2021·陜西·咸陽市實驗中學高二階段練習（理））已知SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項積，且SKIPIF1<0，SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項和，滿足SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）.(1)求證：數(shù)列SKIPIF1<0是等差數(shù)列；(2)求SKIPIF1<0的通項公式；3．（2022·河南安陽·高三期中（理））已知數(shù)列SKIPIF1<0的各項均不為0，其前SKIPIF1<0項的乘積SKIPIF1<0.(1)若SKIPIF1<0為常數(shù)列，求這個常數(shù)；(2)若SKIPIF1<0，設SKIPIF1<0，求數(shù)列SKIPIF1<0的通項公式.4．（2022·江蘇南京·模擬預測）已知數(shù)列SKIPIF1<0的前n項積為SKIPIF1<0，且滿足a1＝1，SKIPIF1<0.(1)求SKIPIF1<0的通項公式；5．（2022·河北邢臺·高三開學考試）數(shù)列SKIPIF1<0的前n項積SKIPIF1<0.數(shù)列SKIPIF1<0的前n項和SKIPIF1<0.(1)求數(shù)列SKIPIF1<0、SKIPIF1<0的通項公式.(2)求數(shù)列SKIPIF1<0的前n項和.6．（2022·全國·高三專題練習）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，各項均為正數(shù)的數(shù)列SKIPIF1<0的前SKIPIF1<0項積為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.(1)求SKIPIF1<0的通項公式；(2)證明：SKIPIF1<0為等比數(shù)列.7．（2022·湖北·模擬預測）已知數(shù)列SKIPIF1<0前SKIPIF1<0項和SKIPIF1<0，SKIPIF1<0的前SKIPIF1<0項之積SKIPIF1<0.(1)求SKIPIF1<0與SKIPIF1<0的通項公式.突破三：累加法1．（2022·安徽·阜陽師范大學附屬中學高三階段練習）已知數(shù)列SKIPIF1<0滿足數(shù)列SKIPIF1<0為等比數(shù)列，SKIPIF1<0，且對任意的SKIPIF1<0．(1)求實數(shù)SKIPIF1<0的值及SKIPIF1<0的通項公式；2．（2022·陜西寶雞·高三期中（文））設數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0.(1)求證：數(shù)列SKIPIF1<0為等差數(shù)列，并求SKIPIF1<0的通項公式；3．（2022·安徽·蕪湖一中模擬預測）已知數(shù)列SKIPIF1<0滿足：SKIPIF1<0．(1)證明數(shù)列SKIPIF1<0為等差數(shù)列，并求數(shù)列SKIPIF1<0的通項公式．4．（2022·云南民族大學附屬中學模擬預測（理））在數(shù)列SKIPIF1<0中，SKIPIF1<0，且對任意的SKIPIF1<0，都有SKIPIF1<0.(1)證明：數(shù)列SKIPIF1<0是等比數(shù)列，并求數(shù)列SKIPIF1<0的通項公式；突破四：累乘法1．（2022·河北張家口·高三期中）已知正項數(shù)列SKIPIF1<0滿足SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項公式；2．（2022·湖南岳陽·高二期中）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且滿足SKIPIF1<0，SKIPIF1<0(1)求數(shù)列SKIPIF1<0的通項公式；3．（2022·福建·高三階段練習）設數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式；4．（2022·湖北·高三階段練習）已知數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，且SKIPIF1<0是公差為2的等差數(shù)列．(1)求SKIPIF1<0的通項公式以及SKIPIF1<0；5．（2022·江蘇泰州·高三期中）設數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)求SKIPIF1<0的通項公式；突破五：構(gòu)造法1．（2022·江蘇鹽城·高三階段練習）已知數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0．(1)求證:SKIPIF1<0是等比數(shù)列,并求SKIPIF1<0的通項公式;2．（2022·重慶八中高三階段練習）記SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項和，已知SKIPIF1<0，SKIPIF1<0是公差為SKIPIF1<0的等差數(shù)列.(1)求SKIPIF1<0的通項公式；3．（2022·河北·三河市第三中學高三階段練習）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0．(1)若SKIPIF1<0，證明：數(shù)列SKIPIF1<0是等比數(shù)列；4．（2022·遼寧·高二階段練習）設數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，若對任意的正整數(shù)n，都有SKIPIF1<0．(1)求SKIPIF1<0的通項公式；5．（2022·黑龍江實驗中學高三期中）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且滿足SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項公式；突破六：倒數(shù)法1．（2022·全國·高三專題練習）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0.求數(shù)列SKIPIF1<0的通項公式;2．（2022·全國·高三專題練習）對負整數(shù)SKIPIF1<0，數(shù)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0依次成等差數(shù)列．(1)求SKIPIF1<0的值；(2)若數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的通項公式；3．（2022·遼寧·昌圖縣第一高級中學高二期末）已知正項數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項公式；4．（2022·黑龍江·龍江縣第一中學高二階段練習）已知數(shù)列SKIPIF1<0的通項公式為SKIPIF1<0，SKIPIF1<0(1)求數(shù)列SKIPIF1<0的通項公式.5．（2022·全國·高三專題練習）已知數(shù)列SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，證明：數(shù)列SKIPIF1<0是等比數(shù)列突破七：隔項等差1．（2022·山西運城·高三期中）已知正項等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式；2．（2022·江蘇鹽城·高三期中）數(shù)列SKIPIF1<0中，SKIPIF1<0．(1)求SKIPIF1<0的通項公式；3．各項均為正數(shù)的數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0．4．（2022·全國·模擬預測）已知數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，SKIPIF1<0．(1)求SKIPIF1<0；突破八：隔項等比1．（2022·山東·肥城市教學研究中心模擬預測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.求數(shù)列SKIPIF1<0的通項公式SKIPIF1<0；第三部分：沖刺重難點特訓1．（2022·河南·民權縣第一高級中學模擬預測（文））已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0．(1)求SKIPIF1<0的通項公式；2．（2022·全國·模擬預測）已知SKIPIF1<0為數(shù)列SKIPIF1<0的前n項和，且SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項公式；3．（2022·江蘇南京·模擬預測）記數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，已知SKIPIF1<0，SKIPIF1<0.(1)求SKIPIF1<0的通項公式；4．（2022·江蘇鹽城·模擬預測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0(SKIPIF1<0)，且SKIPIF1<0(SKIPIF1<0)．(1)求數(shù)列SKIPIF1<0的通項公式；5．（2022·江蘇南通·模擬預測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.(1)證明：數(shù)列SKIPIF1<0是等比數(shù)列；(2)求數(shù)列SKIPIF1<0的通項公式.6．（2022·浙江紹興·一模）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0.有以下三個條件：①SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）；②SKIPIF1<0；③S