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专题1-8数列求和14类题型一网打尽数列求和常见题型梳理【题型1】错位相减【题型2】裂项相消(常规)【题型3】分组求和【题型4】裂项相消(进阶)【题型5】并项求和【题型6】倒序相加【题型7】S2n与S2n-1下标的讨论和处理【题型8】通项含有(-1)n的类型【题型9】奇偶数列求和【题型10】
隔项数列求和(一般并项求和)【题型11】和为等比数列求和【题型12】插入新数列混合求和【题型13】通项含绝对值的数列求和【题型14】取整数列求和数列求和常见题型梳理一、错位相减法类型一:nnncab=(其中na是
等差数列,nb是等比数列)类型二:nnnacb=(其中na是等差数列,nb是等比数列)二、裂项相消法类型一:等差型①)11(1)(1knnkknn+−=+;②1111()(1)(1)211knknknkn=−−+−+类型二:无理型)(11nknknkn−+=+
+类型三:指数型11(1)11()()nnnnnaaakakakak++−=−++++裂项相消进阶1、裂项相加:(-1)n例:()()()21111111nnnnnnn+−=−+++,本类模型典型标志在通项中含有(1)n−乘以一个分式.对于11(1)nnnnnnaaba
a++=−+可以裂项为1111(1)(11)nnnnnnnnnaabaaaa++++−+=−=2、等差数列相邻2两项之积构成的的新数列例如:1(1)(1)(2)(1)(1)3nnnnnnnn+=++−−+一
般式,当公差为k时:1()()(2)()()3knknkknknkknkknkknknkk+=++−−+3、一次乘指数型:分母为一次函数和指数函数相乘例子:122(1)21111(1)2(1)2122(1)2−++−==−=−++++nnnnnnnnnnnn
nnnn一般结构()()()()()11111nnnaknaknabbaknbknbakknbab=−−−++++−++−三、分组求和法3.1如果一个数列可写成nnncab=的形式,而数列na,nb是等差数列或等比数列
或可转化为能够求和的数列,那么可用分组求和法.3.2如果一个数列可写成nnnancbn=为奇数为偶数的形式,在求和时可以使用分组求和法.四、倒序相加法即如果一个数列的前n项中,距首末两项“等距离”的两项之和都相等,
则可使用倒序相加法求数列的前n项和【题型1】错位相减1.已知21nan=−,若数列nb满足11122(23)26nnnabababn++++=−+,求和:121121nnnnnTabababab−−=++++.【答案】13246nnTn+=−−【详解】因为11
122(23)26nnnabababn++++=−+,所以112211(25)26(2)nnnabababnn−−+++=−+,两式相减得1(23)26(25)26(2)nnnnabnnn+=−+−−−(21)2(2),nnn=−又1
12ab=满足上式,所以*(21)2(N),nnnabnn=−又21nan=−,所以2.nnb=则121121nnnnnTabababab−−=++++12123252(21)2nnnn−−=++++−,112
2123252(21)2nnnnTn+−=++++−,两式相减得:1132222(21)2nnnnTn++=++++−−1118(12)2(21)2324612nnnnn−++−=+−−=−−−.2.记数列na的前n项和为nT,且111,(2)n
naaTn−==.(1)求数列na的通项公式;(2)设m为整数,且对任意*nN,1212nnmaaa+++,求m的最小值.【答案】(1)21,1,2,2.nnnan−==;(2)7【分析】(1)由数列na与n
T的关系可得()122nnaan+=,再结合等比数列的通项可得解;(2)利用错位相减法求出1212nnaaa+++,结合范围即可得解.【详解】(1)因为111,(2)nnaaTn−==,所以211aa==,当2n时,112
nnnnnaTTaa+−+===,故()222222nnnaan−−==,且11a=不满足上式,故数列na的通项公式为21,1,2,2.nnnan−==(2)设1212nnnSaaa=+++,
则11S=,当2n时,102122322nnSn−−=++++,故112112232222nnSn−−−=++++,于是()122115222222nnnSn−−−−=++++−()121121252212nnn−−−−−=+−−.整理可得27(2)2nnSn−
=−+,所以7nS,又54968S=,所以符合题设条件的m的最小值为7差比数列的其它处理方式(待定系数法)3.已知()252nnan=−,求nS.【答案】()()()()125212222nnnnn
annnn+=−=++−+=++,22259==+=−=−()()12192292nnnann+=+−−−,()()1121921427214nnnSnn
++=+−+=−+.【题型2】裂项相消(常规)4.已知,证明:.【分析】由,得到,结合裂项求和及,即可得证.【详解】解:由,则.所以.因为,所以,即.5.已知21nan=−,数列na前n项和nS,记21nnnnb
SS++=,设数列nb的前n项和为nT,求证516nT1nnan=+3121234nnaaanaaa+++++1nnan=+1111122nnaann+=+−+11012nn+++1nnan=+()()()()2112111112222nnnnnaa
nnnnnn++++===+−+++3121211111111111232435112nnaaanaaannnn++++=+−+−+−++−+−−++
11113111122124212nnnnnn=++−−=+−+++++11012nn+++3111342124nnnn+−++++3121234nnaaanaaa+++++【
解答】()21212nnnSn+−==,22221111(2)4(2)nnbnnnn+==−++12111413b=−,222111424b=−,322111435b=−,L,1221114(1)
(1)nbnn−=−−+,221111155144(1)(2)4416nTnn=+−−=++.6.已知正项数列na的前n项和为nS,且满足112nnnSaa=+
,(1)求nS,(2)求12233411111nnSSSSSSSS+++++++++【答案】(1)=nSn;(2)11n+−【分析】(1)先令1n=求出首项,再由数列的递推公式,当2n时,1nnnaSS−=−代入并结合等差数列的定义和通项公式求出nS.(2)由第
一问nS的公式,正好利用分母有理化进行化简抵消即可得出结果【详解】(1)根据题意可得0na,当1n=时,1111112Saaa==+,解得11a=,由1nnnaSS−=−,2n代入得11112nnnnnSSS
SS−−=−+−,整理后得111nnnnSSSS−−+=−,即2211nnSS−−=,根据等差数列的定义可知,数列2nS是首项为1,公差为1的等差数列,则21(1)1nSnn=+−=
,=nSn(2)由(1)可知=nSn,12233411111nnSSSSSSSS+++++=++++11111223341nn++++=+++++2132431nn−+−+−+++−11n=+−,1223341111111nnnSSS
SSSSS+++++=+−++++7.已知21nna=−,设1221nnnnbnaa+=−+,求数列nb的前n项和nS.【答案】211121nnSn+=−−−【详解】()()()()()()11121212212121212121nnnnnnnnbnn+++−−−=−+=−+−−−−1
11212121nnn+=−−+−−,所以12nnSbbb=+++2231111111()[(1)(3)(21)]212121212121nnn+=−+−++−+−+−++−+−−−−−−11
(112)1212nnn+−+−=−+−211121nn+=−−−对式子变形后再裂项:一般是分离常数8.已知121nan=−,设214nnncnaa+=,求数列nc的前n项和nT.【解析】()()()()2222122444111111411212141412121
22121nnnnnncnaannnnnnnn+−+=====+=+−−+−−−+−+12311111112335212121nnnTccccnnnnn=++++=+−+−++−=+−++.9.已知24nan=
+,记1nnbna=,数列nb的前n项和为nT,求nT.【解析】()111112442nnbnannnn===−++,∴11111111111432435462nTnn=
−+−+−+−++−+111114212nn=+−−++32384(1)(2)nnn+=−++.10.已知()()*1N1nannn=+,若()221nnbna=+,求
数列nb的前n项和nT.【解析】22222111(1)(1)nnbnnnn+==−++,则()2222222211111111223(1)(1)(1)nnnTnnnn+=−+−++−=−=
+++.11.已知1nnan=+,证明:.【分析】由,得到,结合裂项求和及,即可得证.3121234nnaaanaaa+++++1nnan=+1111122nnaann+=+−+11012nn+++【详解】解:由,则.所以.因为,所以,即【题型3
】分组求和12.已知21nan=−,若数列nb满足,2,nnnanbn=为奇数为偶数,求数列nb的前2n项和2nT.【答案】124423nnn+−−+【详解】因为,2,nnnanbn=为奇数为偶数,所以21
,2,nnnnbn−=为奇数为偶数,所以()()()24221252432nnTn=+++++−+()()2421543222nn=+++−++++()()2122141434422143nnnnnn+−+−−=+=−+−.13.已知2n
na=,设2,log,nnnanban=为偶数为奇数,数列nb的前2n项和为2nT,求2nT.【答案】12214433nnTn+=+−【详解】()()21221321242nnnnTbbbbbbbbb−=+++
=+++++++()242135(21)222nn=++++−++++()()414121214nnn−+−=+−1214433nn+=+−.14.已知2nna=,设mb为数列na在区间(()*0,mmN中的项的个数,
求数列mb前100项的和.【答案】4801nnan=+()()()()2112111112222nnnnnaannnnnn++++===+−+++3121211111111111232435112nnaaanaaannnn++++=+−+−+−+
+−+−−++11113111122124212nnnnnn=++−−=+−+++++11012nn+++3111342124nnnn+−++++3121234nna
aanaaa+++++【详解】由mb为数列na在区间(()*0,mmN中的项的个数,可知10b=,231bb==,45672bbbb====.当815m时,3mb=;当1631m时,4
mb=;当3263m时,5mb=;当64100m时,6mb=.∴12310001122438416532637480bbbb++++=++++++=.15.已知数列na的前n项和nS,且1131,1n
nSSa+=+=,数列nb满足()111,1nnbnbnb+=+=,其中*nN.(1)求na和nb的通项公式;(2)设()3log,4,2nnnancnnb=+为奇数为偶数,求数列nc的前20项和20T.【答案】(1
)13nna−=,nbn=(2)20100011T=【分析】(1)根据11,1,2nnnSnaSSn−==−、累乘法求得na和nb的通项公式;(2)结合分组求和法、裂项相消求和法求得20T.【详解】
(1)对于1131,1nnSSa+=+=,当1n=时,1212131,213aaaaa+=+=+=,当2n时,由131nnSS+=+得131nnSS−=+,两式相减得()132nnaan+=,由于213aa=,所以na是首项为1,公比为3的等比数列,所以13nna−=.对于()111
,1nnbnbnb+=+=,11nnbnbn++=,所以1321122113211221nnnnnbbbbnnbbnbbbbnn−−−−===−−,1b也符合上式,所以nbn=.(2)当n为奇数时,3log1nncan==−;1190,18cc==,所以1319018
10902ccc++++==.当n为偶数时,()()44112222nnnbnnncn==−+++=;所以2420ccc+++111111224462022=−+−++−111102122
21111=−=−=.所以20101000901111T=+=.【题型4】裂项相消(进阶)1、裂项相加:(-1)n例:()()()21111111nnnnnnn+−=−+++,本类模型典型标志在通项中含有(1)
n−乘以一个分式.对于11(1)nnnnnnaabaa++=−+可以裂项为1111(1)(11)nnnnnnnnnaabaaaa++++−+=−=2、等差数列相邻2两项之积构成的的新数列例如:
1(1)(1)(2)(1)(1)3nnnnnnnn+=++−−+一般式,当公差为k时:1()()(2)()()3knknkknknkknkknkknknkk+=++−−+3、一次乘指数型:分母为一次函数和指数函数相乘例子
:122(1)21111(1)2(1)2122(1)2−++−==−=−++++nnnnnnnnnnnnnnnn一般结构()()()()()11111nnnaknaknabbaknbknbakknbab=−−−++++−++−16.若21nan=
−,数列nb满足11(1)nnnnnbaa++−=,nb的前n项和为nT,求nT【答案】11(1)1421nnTn+−=++.【详解】由题可得1111(1)(1)(1)11(21)(21)42121nnnnnn
nnbaannnn++++−−−===+−+−+,所以1111111(1)111(1)114343542121421nnnTnnn++−−=+−++++=+
−++.17.已知()()612nann=++,若()()231nnnbna=+−,求nb的前n项和nT.【详解】()()()112316112nnnnbnann=+−=−+++,所以()12111
1116661233412nnnTbbbnn=+++=−+++++−+++()6312nn=−+−+.18.已知21nan=−,1nnnbaa+=,求数列{nb}的前n项和nT【答案】21(461)3n
Tnnn=+−.【详解】当2n时,()121112116nnnnnnnnnnnnaaaaaaaaaaaa++−+++−+−=−=6nb=,因此,12111()6nnnnnnnbaaaaaa++−+=−,1212312211()(15)66nknnnnnnkbaaa
aaaaaa++++==−=−,则21121221151(15)(21)(21)(23)3(461)6623nnknnnkTbbaaaaannnnnn++==+=−+=−++−+=+−,13b=满足上式,所以21(461)3nTnnn=+−.19.已知21nan=−,()11nnnnbaa+=
−,求数列{nb}的前n项和nT.【答案】2*2*22,2,N221,21,NnnnnkkTnnnkk+==−−+=−【详解】当n为偶数时,()1223344511nnnnTaaaaaaaaaa+=−+−++−+()()()21343511nnnaaaaa
aaaa−+=−++−+++−+()()()24321244212nnnaaann+−=+++==+当n为奇数时,当1n=时,13T=−当3n时,11nnnnTaaT−+=−()()()213232421212212nnnnnn−+−=−−+=−−+经检验,1T也满足上式
,所以当n为奇数时,2221nTnn=−−+综上,数列nb的前n项和2*2*22,2,N221,21,NnnnnkkTnnnkk+==−−+=−20.已知21nbn=−,设()()121(1),
11nnnnnncTbb++=−++为数列nc的前n项和,证明:216nT−.【详解】12121111(1)(1)(1),(1)(1)4(1)41nnnnnnnncbbnnnn+++=−=−=−+++++所以211111111111....1,412232122
21421nTnnnnn=−−++−−−+=−−++由于2111(11)421nTntn=−−+是递减的,所以211111.4216nTT=−=−+21.已知3nnb=,若()()*24141nnn
bcnn+=−N,求数列nc的前n项和【详解】由()()*24141nnnbcnn+=−N,可得()()()()1441132121321321nnnnncnnnn−+==−−+−+,则数列
nc的前n项和为()()0112111111131333335321321nnnn−−+−++−−+()11321nn=−+.已知41nan=−,()1821nnnnnbaa++=−,求数列nb的前21n+项和21nT+.【答案】8102421nn
+−+【详解】()1821nnnnnbaa++=−()()()8214143+=−−+nnnn()1114143=−+−+nnn,所以21123221++=+++++nnnTbbbbb1111111137
71111158183=−+++−++++−+nn118387−+++nn11387=−−+n8102421+=−+nn.22.已知12nna+=,记22(1)nnnabnn++=+,nT为数列nb的前n项和,求n
T.【解析】因为22(1)nnnabnn++=+,所以()()()()1222112122121nnnnnnnbnnnnnna+++===−++++,所以数列nb的前n项和为:()12231111111212222232212nn
nTnn+=−+−++−+()()1111121121212nnnn+=−=−++.23.已知nan=,设14122nnnannnabaaa++++=,证明:1214nbb
b+++.【详解】解:因为14111241422(1)(2)12(2)nnnannnnnannbaaannnnnn++++++++===++++,1112(1)2(1)(2)nnnbnnnn+=−+++,故12nbbb
+++=22311111112122232232342(1)2(1)(2)nnnnnn+−+−++−+++111142(1)(2)4nnn+=−++.【题型5】并项求和一般来说,并项求和的计算量比分组求和要小24.已知21nan=−,
若2πcos3nnnba=,求数列nb的前31n+项和31nT+.【解析】()2π2πcos21cos33nnnnban==−,【法一】并项求和()()()()()3133162π62π6π63cos6
1cos61cos333nnnnnnbbbnnn−+−+++=−+−++()()()22π63cosπ61cos2π61cos33nnn−=−+−++化简得()()3133111636161022nnnbbb
nnn−+++=−−+−−+=,故()()()311234567313311102nnnnTbbbbbbbbbbb+−+=++++++++++=+=−【法二】分组求和311233231331nnnnnTb
bbbbbb+−−+=+++++++()()()311111135165636112222nTnnn+=−+−+++−−+−−+−()1612n++−()()()()1
65363561111161222222nnnnnnn+−+−+−=−+−+++−22233113232222nnnnnn=−+−++−−=−,所以,数列nb的前31
n+项和3112nT+=−25.(2023秋·湖南长郡中学校考)已知nS是数列na的前n项和,1232,3,4aaa===,数列12nnnaaa++++是公差为1的等差数列,则40S=.【答案】366【分析】设12nn
nnbaaa++=++,易得()9118nbnn=+−=+,再由4012538Sabbb=++++求解.【详解】解:设12nnnnbaaa++=++,由题意知nb是公差为1的等差数列,则11239baaa=++=,故()9118nbnn=+−=+,则
21110bb=+=,故()()()()25381323828583881383642bbb++++=++++++=+=.于是()()()401234567383940Saaaaaaaaaa=+++++++++,12
5382364366abbb=++++=+=.26.已知21nan=−,记()1nnnbS=−,求数列nb的前30项的和30T.【解析】2(121)2nnnSn+−==,所以2(1)(1)=−−=nnn
nbnS,所以2222223012342930T=−+−++−+()()()()()()2112433430292930=−++−+++−+12342930=++++++30(130)4652+==27.已知212nna−=,设11b=,1,,nnna
nbbnn+=−+为奇数为偶数,求数列nb的前2n项和2nS.【详解】当n为奇数时,2112nnnba−+==;则当n为偶数时,1nnbbn++=.2122nnSbbb=+++()()()()1223452221nnnbbbbbbbb−−
=+++++++()2112422nan−=+++++−()()43432222112212nnnnnn−−+−−=++=+−+.【题型6】倒序相加28.“数学王子”高斯是近代数学奠基者之一,他的数学研究几乎遍及所有领域,并且高斯研究出很多数学理论,比如高斯函数、倒序相加法、最小二乘
法、每一个n阶代数方程必有n个复数解等.若函数()22log1xfxx=−,设()112311,,2nnaaffffnnnnnn−==++++N,则1210aaa+++=.【答案】46【分析】先证()()12fxfx+−
=,由倒序相加法可得通项,然后可解.【详解】因为函数()22log1xfxx=−的定义域为()0,1,设()()1122,,,MxyNxy是函数()yfx=图象上的两点,其中()12,0,1xx,且121xx=+,则有()()()12121212222121212
224logloglog2111xxxxyyfxfxxxxxxx+=+=+==−−−++,从而当()0,1x时,有:()()12fxfx+−=,当2n时,1231nnaffffnnnn−
=++++L,121nnnafffnnn−−=+++,相加得112nnaffnn−=++221122nnf
fffnnnnn−−++++=−所以1nan=−,又11a=,所以对一切正整数n,有1,11,2nnann==−;故有12:a
a++()101123946a+=+++++=.29.(2023·江西南昌·统考三模)“数学王子”高斯是近代数学奠基者之一,他的数学研究几乎遍及所有领域,在数论、代数学、非欧几何、复变函数和微分几何等方面都作出了开创性的贡献.我们高中阶段也学习过很多高斯的数学理论,比如高斯函数、
倒序相加法、最小二乘法等等.已知某数列的通项251,262521,26nnnann−=−=,则1251...aaa+++=()A.48B.49C.50D.51【答案】D【分析】分离常数后可得522nnaa−+=,再利用倒序相加法,即可求解.【详解】当26n时,
2512521112522522(26)nnnannn−−+===+−−−,52111122(26)2(5226)nnaann−+=+++=−−−,1251Saaa=+++,51491Saaa=+++,1512495112()()()251Saa
aaaa=++++++=,51S=,即1251...51aaa+++=.【题型7】S2n与S2n-1下标的讨论和处理30.已知数列()()21,2,nnnnan+=为奇数为偶数(1)求数列na的前20项和20T(2)
求数列na的前2n项和2nT.(3)求数列na的前21n−项和21nT−.(4)求数列na的前n项和nT【详解】(1)201351924620()()TCCcCccCc=+++++++++24620
(371139)(2222)=+++++++++()21011214(330)1062642143−++=+=−(2)由(1)知()()21,2,nnnncn+=为奇数为偶数,数列{}nc的奇数项是首项为3,公差为
4的等差数列,偶数项是以首项为4,公比为4的等比数列.记13521nkAcccc−=++++,2462nkBcccc=++++2141kck−=−,故234122nkAkkk+−==+222kkc=,故()224124241433kknB−==−−122412
3kkTABkk+−=+=++(3)2212124123kkkkTTckk−−−=−=++(4)当n为偶数时,记n=2k则有1224123kkTkk+−=++,故222123nnnnT++−=+当n为奇数时,记n
=2k-1则有2214123kkTkk−−=++,故21322423nnTnn+++=+−故()22122212,22324,2,133nnnnnnkTnnnkkN+++++−+−+===−+31.已知21,nan−=2na=212n+,记na的前n项和为nS
,2023nS,求n的最小值.【分析】解法一:枚举;解法二:分组求和得出()()2841123kkkkS−+=+,进而得出()21124823kkkkS−+−=+,求解即可得出答案;解法三:分组求和得出()21124823kkkk
S−+−=+,求解即可得出答案.【详解】解法一:9123456789Saaaaaaaaa=++++++++()()135792468aaaaaaaaa=++++++++()()3579123452222156806952023=++++++++=+=,又11109106952
27432023SSa=+=+=;又0na,则1nnSS+,且9102023SS,所以n的最小值为10.解法二:*kN时,21232kkSaaaa=++++()()135212462kkaaaaaaaa−=
+++++++++()()357211232222kk+=+++++++++()()841123kkk−+=+,()()()2121228411124822323kkkkkkkkkkSSa+−−++−=−=+−=+,
所以5925156248695202323SS−−==+=,()51025841562743202323SS−==+=,又0na,则1nnSS+,且9102023SS,所以n的最小值为10.解法三:当*kN时,2112321kkSaaaa−−=++++()
()1352124622kkaaaaaaaa−−=+++++++++()()357211232222kk−=+++++++++()()()118411114821423kkkkkk−−−++−=+=+−,所以()4925184156695202323SS−−==+=,251
10910695227432023SSa+=+=+=.又0na,则1nnSS+,且9102023SS,所以n的最小值为10.32.(2023·湖南岳阳·统考三模)已知3nna=,若13log
,,nnnanban=为奇数为偶数,求数列nb的前n项和nT.【解析】,3,nnnnbn−=为奇数为偶数,当n为偶数时,()()1213124nnnnTbbbbbbbbb−=+++=+++++++()()24131333nn=−++
+−++++()2919112219nnn−+−=−+−()293184nn=−−;当n为奇数时()()211111931384nnnnnnTTb+++++=−=−−−()211193884nn++=−−;综上所述:()()212
1193,884931,84nnnnnTnn++−−=−−为奇数为偶数.【题型8】通项含有(-1)n的类型33.已知()*nann=N,若()21nnnba=−,求数列nb的前n项和
nT.解题思路点拨:代入得:注意到通项中含有“”,会影响最后一项取“正还是负”,通过讨论的奇偶,结合分组求和.奇偶项通项不同,采用分组求和可作为一个解题技巧(注意到本例求解的,代入最后一项,是正,还是负,需要讨论)(讨论时优先讨论为偶数)为奇数
为偶数当为奇数时,为偶数,即:注意到为偶数,所以可使用偶数项和的结论,代入左侧求和结果:,则:,整理:综上:34.已知nan=,设数列()()24141nnnabnn+=−−N,数列化nb的前2n项和为2nT35.在数列{an}中,若()1121nnnaan++−−=,则数列{an}的前1
2项和等于_________.【答案】78因为()1121nnnaan++−=−,所以211aa−=,323aa+=,435aa−=,547aa+=,659aa−=,76aa+=11,87aa−=13,98aa+=15,1
09aa−=17,1110aa+=19,1211aa−=21.从第一个式子开始,相邻的两个式子作差得:1357911aaaaaa+++===2.从第二个式子开始,相邻的两个式子相加得:42681012aaaaaa+++=8,=24,=40,把以上的式子依次相加可得:1
2121112Saaaa=++++()()()()()()135791124681012aaaaaaaaaaaa=+++++++++++22282440+=++++=78.36.已知13nna+=,若()31lognnnba=−,求数列nb的前n项和nT.思路点拨:,注意到通项
中含有“”,会影响最后一项取“正还是负”,通过讨论的奇偶,结合分组求和.奇偶项通项不同,采用分组求和可作为一个解题技巧(注意到本例求解的为偶数项和,代入最后一项,一定是正,故不需要讨论)分组求和【答案】,23,2
nnnTnn=+−为偶数为奇数【详解】()()()31log11nnnnban=−=−+故当n为偶数时,()()()234512nnTnn=−++−+++−++=当n为奇数时,()()()()()123451112nnTnnnn−=−++−
+++−−+−+=−+32n+=−,所以,23,2nnnTnn为偶数为奇数=+−37.已知数列na中,()112,1nnnanaaa+=−=+.(1)求证:数列1nan+是常数
数列;(2)令(1),nnnnbaS=−为数列nb的前n项和,求使得99nS−的n的最小值.【答案】(1)证明见解析;(2)最小值为67.(1)由()11nnnnaaa+−=+得:()111nnnana+=++,即()1111nnaan
nnn+=+++11111nnaannnn+=+−++,即有111,1nnaann+++=+数列1nan+是常数数列;(2)由(1)知:()1113,31,(1)31nnnnaaanbnn+=+==−=−−即()31,31,nnnbn
n−=−−为偶数为奇数,当n为偶数时,()()()()32581134312nnSnn=−++−+++−−+−=,显然99nS−„无解;当n为奇数时,()()11313131122nnnnnSSan++++=−=−+−=−,令99n
S−,解得:66n…,结合n为奇数得:n的最小值为67.38.已知数列{}na的前n项和nS,11a=,0na,141nnnaaS+=−.(1)计算2a的值,求{}na的通项公式;(2)设1(1)nnnnbaa+=−,求数列{}nb的前2n项和2nT.
【答案】(1)23a=,21nan=−(2)24(21)nTnn=+【分析】(1)根据11,1,2nnnSnaSSn−==−,作差得到24nnaa+−=,再根据等差数列通项公式计算可得;(2)由(1)可得(1)(21)(21)nnbnn=−−+,利用并项求和法计算
可得;【详解】(1)解:当1n=时,12141aaa=−,解得23a=,由题知141nnnaaS+=−①,12141nnnaaS+++=−②,由②−①得121()4nnnnaaaa+++−=,因为0na
,所以24nnaa+−=,于是:数列{}na的奇数项是以11a=为首项,以4为公差的等差数列,即()2114(1)432211nannn−=+−=−=−−,偶数项是以23a=为首项,以4为公差的等差数列,即234(1)41nann=+−=−所以{}na的通项公式21nan=
−;(2)解:由(1)可得(1)(21)(21)nnbnn=−−+,212(43)(41)(41)(41)4(41)nnbbnnnnn−=−−−+−+=−+21234212(341)()()()4[37(41)]44(21)2nnnnnTbb
bbbbnnn−+−=++++++=+++−==+.39.已知1=−−nann,()1nnnba=−,求{nb}的前64项和64T.【答案】nb的前64项和648T=.【详解】(1)(1)(1)(1)1nnnnnbnnann−−===−+−−−,641(
21)(32)(641642)(64641)T=−++−++−−+−++−=1122362636364−++−−+−−++648==,即:648T=.∴nb的前64项和648T=.【题型9】奇偶数列求和重庆一中
月考40.已知数列na满足12a=,11,,2,.nnnanaan++=为奇数为偶数若()*123nnSaaaan=++++N,求2nS.【答案】3238nn+−−【详解】证明:11,,2,
.nnnanaan++=为奇数为偶数,设21nnba=+()()122212122111222212nnnnnnnbaaaaab++++=+=++=+=+=+=,又()12111140baa=+=++=
12nnbb+=,nb为以4为首项,2为公比的等比数列.1121422nnnnba−+=+==,1221nna+=−,又1221121nnnaa+−=+=−,12122nna+−=−,所以()()2135212462nnnSaaaaaaaa−=++
+++++++()()341241222381212nnnnnn+−−=−+−=−−−−.2021·新高考1卷T1741.已知数列na满足11a=,11,2,nnna
naan++=+为奇数为偶数,求na的前20项和.【答案】300.【详解】21222212,1nnnnaaaa+++=+=+,设2nnba=所以2223nnaa+=+,即13nnbb+=+,且12
1+12baa===,所以nb是以2为首项,3为公差的等差数列,于是122,5,31nbbbn===−20123201351924620++++++++()()Saaaaaaaaaaaa=+=+++1231012310(1111)bbbbbbbb=−+−+−++−+
++++110()102103002bb+=−=.42.(广东实验中学校考)已知数列满足,且的前100项和(1)求的首项;(2)记,数列的前项和为,求证:.【答案】(1)(2)证明见解析【分析】(1)分为奇数和为偶数两种情况进而讨论即可求解;(2)结合(1)的结论,利用裂项相消法即可求解.【
详解】(1)当为奇数时,;则偶数项构成以为公差的等差数列,所以当为偶数时,;当为偶数时,,则奇数项构成以1为公差的等差数列,所以当为奇数时,,则,又,na()121,13,2nnnanaan+−=
+当为奇数时当为偶数时na1003775S=na1a2121nnnbaa−=nbnnT32nT11a=nnn()11112123122nnnnaaaa+−−=−=+−=+2n22naan=+−n()()111113213122nnnnaaaa+−−=+=−+=+
n112nnaa−=+121,22,nnanaann−+=+−当为奇数时当为偶数时2121aa=−所以,解得,.(2)由(1)得,,,,当时,,∴,综上,知.【题型10】隔项数列求和(一般并项求和)43.已知数列{}na满足11a=,14nnaan++=,则
100S=________【答案】10000【详解】数列na满足11a=,14nnaan++=,因为11a=,14nnaan++=,所以,()()()100123499100Saaaaaa=++++++(
)()4199504143499413599100002+=+++=++++==44.若数列na的前n项和为nS,且12nnnaa++=,则10S=()A.684B.682C.342D.341【答案】B【分析】根据等比数列求和公式以及并项求和法得出结
果.【详解】1212aa+=,3432aa+=,5652aa+=,7872aa+=,91092aa+=,所以13510222S=+++5792(14)2268214−+==−.45.(深圳一模)记nS,为数列
na的前n项和,已知142nnaan−+=−,14a=求nS.【答案】22,2,nnnnSnnn+=++当为偶数时当为奇数时【详解】解:()141242nnaann−+=−+=−,*nN,2n.当n为偶数时,()()
()()1234164222nnnnnSaaaaaa−+−=++++++=2nn=+;当n为奇数时,()()()()12345111042242nnnnnSaaaaaaa−−+−=+++++++=+22nn=++.()()10013992
4100115036253775Saaaaaaa=+++++++=+=11a=21nan−=221nan=−11312Tb==2n()()212111111212221nnnbaann
nnnn−===−−−−121111113131122231222nnTbbbnnn=++++−+−++−=−−32nT综上所述,22,2,nnnnSnnn+=++当为偶数时当为奇数时【题型11】和为
等比数列求和46.已知数列{}nb中,*1111,2,nnnbbbnN−+=+=,求数列2{}nb的前n和.【答案】2224.939nn−−47.已知数列na满足2123nnnaaa++=+,112a=,232a=.(1)求na的通项公式.思路点拨:根据题意:,可推出,两式作差变
换下标,写成所以,,.......累加,得累加求通项所以数列的前n和为求和(2)若数列na的前n项和为nS,且()*127N4nSnn+−恒成立,求实数的取值范围.【答案】(1)1132nna−=,(2)481【分析】(1)将2123nnnaaa++=
+两边同时加1na+,结合等比数列的定义证明可得1123nnnaa−++=,再构造数列1132nna−−,求解首项分析即可;(2)根据等比数列的前n项公式可得()1314nnS=−,参变分离可得()*27N43nnn−,再根据273nn−的单调性求解最大值即可.【详解】(
1)由2123nnnaaa++=+可得()2111333nnnnnnaaaaaa+++++=+=+,且2113222aa+=+=,故1nnaa++是以2为首项,3为公比的等比数列,故1123nnnaa−++=,所以11113322nnnnaa−+−=−−,又1111302a
−−=,故1211111333....0222nnnnnnaaa−−+−−=−−=−==,即1132nna−=.(2)由(1)1132nna−=为等比数列,故()()1131231134nnnS−==−−,故()*127N4nSnn+
−即()*27N43nnn−恒成立,求273nn−的最大值即可.设()*27,N3nnnbn−=,则()11112562125271643333nnnnnnnnnnnbb++++−−−−−−−=−==,令116403nn+−有4n,故当4n时,nb随n的增大而增
大;当5n时,nb随n的增大而减小.又540bb−=,故54bb=为()*27,N3nnnbn−=的最大值,为42471381−=,所以1481,4812023·杭州二模48.设公差不为0的等差数列na的前n项和为nS,520S=,2325aaa
=.(1)求数列na的通项公式;(2)若数列nb满足11b=,1(2)nannbb++=,求数列2nb的前n项和nT.【答案】(1)22nan=−,Nn+;(2)24133nn−−【分析】(1)根据等差数列性质设出公差和首项,代入题中式
子求解即可;(2)列出1nnbb++通项公式,根据通项求出1nnbb++的前n项和,再根据通项求出nb的前2n项和,两式相减解得2nb的通项公式,最后分组求和求出数列2nb的前n项和nT.【详解】(1)5335204Saa
===,设公差为d,首项为1a()()223233352322adaaaaadadd=−+=+−=,因为公差不为0,所以解得2d=,311240aada=+==,数列na的通项公式为22nan=−,
Nn+.(2)12212)2(2)(nannnnbb−+−==+=()()()()123456212nnbbbbbbbb−++++++++①024222222n−=++++()11414n−=−4133n=−()()(
)()122334212nnbbbbbbbb−++++++++②012222222n−=++++()2111212n−−=−2121n−=−2①-②得211241=22133nnnbb−+−−+
,解得21222=4233nnnb−−−()()81421428414122413214143333333nnnnnnTnnn−−−−−=−−=−−=−−−【题型12】插入新数列混合求和49.已知等差数列na
的首项11a=,公差10d=,在na中每相邻两项之间都插入4个数,使它们和原数列的数一起构成一个新的等差数列nb,则2023b=()A.4043B.4044C.4045D.4046【答案】C【分析】根据等差数列的
性质求出21nbn=−,再代入即可.【详解】设数列nb的公差为1d,由题意可知,11ba=,62ba=,61211105bbaad−=−==,故12d=,故()12121nbnn=+−=−,则20232023214045b=−=50.已知()3
1nan=−对所有正整数m,若12mkkaa+<<,则在ak和ak+1两项中插入2m,由此得到一个新数列{bn},求{bn}的前40项和.【答案】1809【分析】考虑到674022a,634992a=,从
而确定{}nb的前40项中有34项来自{}na,其他6项由2k组成,由此分组求和.【详解】由40117a=.所以674022a,又634992a=,所以nb前40项中有34项来自na.故()()12612401234222bbbaaa+++=+++
++++()()61342213416831261809221aa−+=+=+=−51.已知数列na的通项公式515nan=+,在数列na的任意相邻两项ka与()11,2,kak+=之间插入2k个4,使它们和原数列的项构成一个新的数列nb,记新数列nb的前n项和为nS,则6
0S的值为.【答案】370【分析】依题意,确定前60项所包含数列na的项,以及中间插入4的数量即可求和.【详解】因为ka与()11,2,kak+=之间插入2k个4,120a=,225a=,330a=,435a=,540a=,其中1a,2a之间插入2个
4,2a,3a之间插入4个4,3a,4a之间插入8个4,4a,5a之间插入16个4,5a,6a之间插入32个4,由于6248163660++++=,624816326860+++++=,故数列n
b的前60项含有na的前5项和55个4,故602025303540455370S=+++++=52.已知数列13−=nna,在na和1na+之间插入n个数,使这2n+个数构成等差数列,记这个等差数列的公差为nd,求数列1nd的前n项和nT.【答案
】11525883nnnT−+=−【详解】解:由题可知1113323111nnnnnnaadnnn−−+−−===+++,得111123nnnd−+=,则012211213141112323232323nnnnnT−−+=+++++L,③12311121314111
32323232323nnnnnT−+=+++++L,④③−④得21211111113233323nnnnT−+=++++−L11111115251331122344313nnn
nn−−++=+−=−−,解得11525883nnnT−+=−.53.已知21nan=−,在na与1na+之间插入一项nb,使na,nb,1na+成等比数列,且公比为()0nnqq,求数列lnnq的前n项和nT.【
答案】()1ln212nTn=+【详解】因为na,nb,1na+成等比数列,且公比为nq,所以21nnnaqa+=因为21nan=−,所以()22121nnqn−=+,即22121nnqn+=−因为0nq,则有:212lnln21nnqn+=−可得:()()
1211lnlnln21ln212212nnqnnn+==+−−−化简可得:()()()1211lnlnlnln3ln1ln5ln3ln21ln21ln2122nnTqqqnnn=+++=−+−+++−−=+所
以数列lnnq的前n项和:()1ln212nTn=+54.已知12nna-=,在数列na中的ia和()*1iai+N之间插入i个数1m,2m,3m,…,im,使ia,1m,2m,3m,…,im,1ia+成等差数列,这样得到一个新数列nb,设数列nb的前n项和为n
T,求21T.【答案】5132.【详解】因为在数列na中的ia和()*1iai+N之间插入i个数,则在列nb的前21项中,就是在1a到6a每两项之间各插入一组数,共插入五组,数列nb的前21项为1122334564,,,,,,,,,,
,amammammma∴1122334562146amammammmTaa=+++++++++++()()()()3456121223344556352222aaaaaaaaaaaaaaaa+++=++++++++++++()()
()()348516321212244828161632222+++=++++++++++++5132=.55.己知数列na满足nan=,在1,nnaa+之间插入n个1,构成数列nb:12341,1,1,1,1,,,,,,1aaaaL,则
数列nb的前100项的和为()A.178B.191C.206D.216【答案】A【详解】数列{}na满足nan=,在na,1na+之间插入n个1,构成数列1{}:nba,1,2a,1,1,3a,1,1,1,4a,所以共有
2(1)(11)1[12(1)]()22nnnnnnn−+−++++−=+=+个数,当13n=时,11314912=,当14n=时,114151052=,由于nan=,所以()100121311313()(10013)1871782Saaa+=++++−=+=.【题型13】通
项含绝对值的数列求和56.已知211nan=−,求数列na的前n项和为nT.【答案】2210,151050,6nnnnTnnn−=−+.【详解】当15n时,0na,即有()2121210nnnnTaaaaaaSnn=+++=−+++=−=−.当6n时,0na
,()()21256551050nnnTaaaaaSSSnn=−++++++=−+−=−+,即有2210,151050,6nnnnTnnn−=−+.57.已知210nan=−,求数列||na的前n项和为nT.【答案】
229,15940,6nnnnTnnn−=−+.【详解】当15n时,0na,即有29nnTSnn=−=−;当6n时,0na,255940nnTSSSnn=−−=−+,即有229,15940,6nnnnTnnn−=−+.58.已知()()1021
10122nnann==−,设nnba=,求数列nb的前n项和nT.【答案】()()22100350100500351nnnnTnnn−++=−+【详解】当50n时,nnba=,所以123nnTaaaa=+++
+39997951012n=+++++−()()991012331002nnnn+−=+=+−,当51n时,nnba=−,123505152nnTaaaaaaa=++++−−−−()5012312nnTaa
aaa−=−+++++()50063100nn=−−−21005003nn=−+.综上所述()()22100350100500351nnnnTnnn−++=−+.【题型14】取整数列求和59.已知数列na满足(
)1121,1nnanaan+=−−=+,记na为不小于na的最小整数,nnba=,则数列nb的前2023项和为()A.2020B.2021C.2022D.2023【答案】A【分析】利用裂项相消求和可得答案.【详解】由题意得11121
nnaann+−=−+,则当2n时,()()()112211nnnnnaaaaaaaa−−−=−+−++−+=1111122221111212nnnnn−+−++−−=−−−−,当1
n=时也满足上式,所以()*21nann=−N,所以1234562211,00,11,11,134bbbbbb=−=−===−==−====,故nb的前2023项和为101112020−+++++=60
.已知2nan=+,2nnb=,设3nnac=,求数列nnbc的前9项的和9S.(注:x表示不超过x的最大整数)【答案】2926【详解】2[][]33nnanc+==,则有123456
7981,2,3ccccccccc=========,依题意,234678995121212222222323232S=++++++++=2926,综上,2nan=+,2nnb=,92926S=.61.已知(2)nna=−,设12nnb+=,求数列nnab的前
10项和10T.([]x表示不超过x的最大整数)【答案】3186.【详解】依题意,123456789101,1,2,2,3,3,4,4,5,5bbbbbbbbbb==========,则2345678910102(2)2(2)2(2)
3(2)3(2)4(2)4(2)5(2)5(2)T=−+−+−+−+−+−+−+−+−+−23456789102(2)2(2)(2)3(2)(2)4(2)(2)5(2)(2)=−+−+−+−+−+−+
−+−+−+−457922324252=++++216965122560=++++3186=.62.(重庆八中月考)已知21nan=−,若x表示不超过x的最大整数,如1,22,2,12−=−=,求22212111n
aaa+++的值.【答案】1【详解】222111(21)441nannn==−−+,当2n时,22111114441nannnn=−−−,故22212111111111111151111412231444naaannn++++−+
−++−=+−+=−,当1n=时,211514a=,所以对任意的*nN,都有2221211154naaa+++,又222212111111naaaa+++=,所以22212111514naaa+++.所以222121111naaa+
++=.