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专题27数列求和【考点预测】一.公式法(1)等差数列na的前n项和11()(1)22++==+nnnaannSnad,推导方法:倒序相加法.(2)等比数列na的前n项和111(1)11,,==−−nnnaqSaqqq,推导方法:乘公比,错
位相减法.(3)一些常见的数列的前n项和:①112123(1)==++++=+nkknnn;122462(1)==++++=+nkknnn②21(21)135(21)=−=++++−=nkknn;③2
2222116123(1)(21)==++++=++nkknnnn;④3333321(1)2123[]=+=++++=nknnkn二.几种数列求和的常用方法(1)分组转化求和法:一个数列的通项公式是由若干个等差或等比或可求和的数列组成的
,则求和时可用分组求和法,分别求和后相加减.(2)裂项相消法:把数列的通项拆成两项之差,在求和时中间的一些项可以相互抵消,从而求得前n项和.(3)错位相减法:如果一个数列的各项是由一个等差数列和一个等比数列的对应项之积构成的,那么求这个数列的前n项和即可用错位相减法求解.(4)倒序
相加法:如果一个数列na与首末两端等“距离”的两项的和相等或等于同一个常数,那么求这个数列的前n项和即可用倒序相加法求解.【方法技巧与总结】常见的裂项技巧积累裂项模型1:等差型(1)111(1)1=−++nnnn(2
)1111()()=−++nnkknnk(3)21111()4122121=−−−+nnn(4)1111(1)(2)2(1)(1)(2)=−+++++nnnnnnn(5)211111()(1)(1)(1)2(1)
(1)==−−−+−+nnnnnnnnn(6)22111414(21)(21)=+−+−nnnn(7)314(1)(3)11114()()(1)(2)(3)(1)(2)(3)2312++−+==−−−
++++++++++nnnnnnnnnnnnn(8)1(1)(1)(2)(1)(1).3+=++−−+nnnnnnnn(9)1(1)(2)(1)(2)(3)(1)(1)(2)4++=+++−−++nnnnnnnnnnn(10)1111(1)(2)(3)3(1
)(2)(1)(2)(3)=−++++++++nnnnnnnnnn(11)2222211111)(()+=−++nnnnn(12)222211112)42)((+=−++nnnnn积累裂项模型2:根式型(1)111=+−++nnn
n(2)11()=+−++nknknkn(3)11(2121)22121=+−−−++nnnn(4)2211(1)11111(1)(1)1++++==+−+++nnnnnnnn(5)333222121121+++−+−+nnnnn3333322233111(2112
1)2+−=+−−+++−+−+=nnnnnnnnn(6)221(1)1(1)111(1)(1)11(1)(1)+−++−+===−++++++−+nnnnnnnnnnnnnnnnnnnn积累裂项模型3:指数型(1)11112(21)(21)11
(21)(21)(21)(21)2121++++−−−==−−−−−−−nnnnnnnnn(2)113111()(31)(31)23131++=−−−−−nnnnn(3)122(1)21111(1)2(1)2122(1)2−++−==−=−++++
nnnnnnnnnnnnnnnn(4)1111(41)31911333(2)2(2)22−+−−−=−=−+++nnnnnnnnnnn(5)11(21)(1)(1)(1)(1)++−−−=−++nnnnnnnn(6)13−=nnan,设
1()3[(1)]3−=+−−+nnnaanbanb,易得11,24==−ab,于是111(21)3(23)344−=−−−nnnann(7)222111(1)2(1)(1)(42)2(1)(42)2(1)2(1)2(1)2+++−++++−++
−++==+++nnnnnnnnnnnnnnnnnnnnnn1111(1)1111(1)(1)(1)()22(1)2222(1)2++++−−−=+−+=−+−++nnnnnnnnnnnnnn积累裂项模型4:
对数型11logloglog++=−nanaaannaaa积累裂项模型5:三角型(1)11(tantan)coscossin()=−−(2)11tan(1)tancoscos(1)sin1=+−+nnnn(3)1tant
an(tantan)1tan()=−−−(4)tantan(1)tantan(1);tan1tan(1)1tantan(1)−−=−=−−=+−nnnannnnn,则tantan(1)tantan(1)
tantan(1)1,1tan1tan1−−−−−=−=−nnnnnnna积累裂项模型6:阶乘(1)1!(1)!1(1)!+=−+nnnn(2)2(2)(2)!(1)!(221111=-!(1)!!(2)!!(2)!2)++++++===++++++nnnnnnnnnnnnn常见放缩公式:
(1)()()21111211=−−−nnnnnn;(2)()2111111=−++nnnnn;(3)2221441124412121==−−−+nnnnn;(4)()()()11!111112!!!11+===−−−−rrnrrnTCrnrnrn
rrrrr;(5)()1111111312231++++++−nnnn;(6)()()1222121==−−++−+nnnnnnnn;(7)()122211==−+++++nnnnnnn;(8)
()122222212111212122===−−+++−++−++nnnnnnnnn;(9)()()()()()()()1211222211212121212122212121−−−===−−−−
−−−−−−nnnnnnnnnnnnn()2n;(10)()()()()32111111111111+−−==+−−−+−+nnnnnnnnnnnnn()()111111121111211++−=−=−
+−−−+−+nnnnnnnnnnn()112211−−+nnn;(11)()()()32212221111==−+−−+−+nnnnnnnnnnnnn()()()2122211−−−==−−−nnnnnnn;(12)()()0121112222111111
1===−−++−+++−nnnnnCCCnnnn;(13)()()()111121122121212121−−−=−−−−−−nnnnnnn.(14)21211112()2()+−+++−−==−
nnnnnnnnn.【题型归纳目录】题型一:通项分析法题型二:公式法题型三:错位相减法题型四:分组求和法题型五:裂项相消法题型六:倒序相加法题型七:并项求和题型八:先放缩后裂项求和题型九:分段数列求和【典
例例题】题型一:通项分析法例1.(2022·全国·高三专题练习)求和()()()22122323322332322nnnnnS−−=++++++++++.【解析】∵122332322nnnnna−−=++++222231333nn=++++
111213332213nnnn+++−==−−,∴()()2312312)4)9(13333222(213nnnnnS+++−=+++
−++=−−+−.2231222nnnS++=−−例2.数列9,99,999,的前n项和为()A.10(101)9nn−+B.101n−C.10(101)9n−D.10(101)9nn−−【解析】解数列通项
101nna=−,23(10101010)nnSn=++++−10(110)110nn−=−−10(101)9nn=−−.故选:D.例3.求数列1,(12)+,2(122)++,,21(1222)n−++++,的前n项之和.【解析】解:由于1212112222121nnnna−−
=++++==−−,所以前n项之和12(21)(21)(21)nnT=−+−++−123(2222)(111)n=++++−+++2(21)21nn−=−−122nn+=−−.【方法技巧与总结】先分析数列通项的特点,再选择合适的方法
求和是求数列的前n项和问题应该强化的意识.题型二:公式法例4.已知等差数列{}na中,29a=,521a=.(1)求{}na的通项公式;(2)令2nanb=,求数列{}nb的前n项和nS.【解析】解:(1)设数列{}na的公差为d,由题意得19214adad+==+解得15a=,4
d=,{}na的通项公式为41nan=+.(2)由41nan=+得412nnb+=,{}nb是首项为512b=,公比42q=的等比数列.54442(21)32(21)2115nnnS−−==−.例5.如
图,从点1(0,0)P做x轴的垂线交曲线xye=于点1(0,1)Q,曲线在1Q点处的切线与x轴交于点2P,再从2P做x轴的垂线交曲线于点2Q,依次重复上述过程得到一系列点:1P,1Q;2P,2Q;nP
,nQ,记kP点的坐标为(kx,0)(1k=,2,,)n.(Ⅰ)试求kx与1kx−的关系(2)kn剟;(Ⅱ)求112233||||||||nnPQPQPQPQ++++.【解析】解:(Ⅰ)设11(kkPx−−,0),由xye=得111(,)kxkkQxe−−−点1kQ−处切线方程为111(
)kkxxkyeexx−−−−=−由0y=得11(2)kkxxkn−=−剟.(Ⅱ)10x=,11kkxx−−=−,得(1)kxk=−−,(1)||kxkkkPQee−−==112233||||||||nnnSPQPQP
QPQ=++++112(1)11111nnneeeeeeee−−−−−−−−−=++++==−−声明:试题解析著作权属所有,未经书面同意,不得复制发布日期:2022/7/1012:47:46;用户:18316341968;邮箱:183163419
68;学号:32362679【方法技巧与总结】针对数列的结构特征,确定数列的类型,符合等差或等比数列时,直接利用等差、等比数列相应公式求解.题型三:错位相减法例6.(2022·全国·高三专题练习)“一尺之棰,日取其半,万世不竭”出自我国古代典籍《庄子·天下》,其中蕴含着
等比数列的相关知识.已知长度为4的线段AB,取AB的中点C,以AC为边作等边三角形(如图①),该等边三角形的面积为1S,在图①中取CB的中点1C,以1CC为边作等边三角形(如图②),图②中所有的等边三角形
的面积之和为2S,以此类推,则3S=___________;1niiiS==___________.【答案】21316;()434413931294nnnn−+++.【解析】依题可知,
各等边三角形的面积形成等比数列,公比为14,首项为3,所以1314431113414nnnS−==−−,即321316S=;11114341431343nnnniiiiiiiiiSii=====−
−=,而()112ninni=+=,设21111124444nnniiiTn===+++,()23111111012144444nnnTnn+=++++−+
,作差得:2113111114144444334nnnnTnn++=+++−=−+,所以4419394nnnT
=−+,所以1niiiS==()434413931294nnnn−+++.故答案为:21316;()434413931294nnnn−+++.例7.(2022·内蒙古·海拉
尔第二中学模拟预测(理))已知数列{}na的前n项和343nnnS−=,记2nnnab=,则数列{}nb的前n项和nT=_______.【答案】2332nn+−【解析】当1n=时,114113aS−===,当2n时,()()3321411444133nn
nnnnnaSSnn−−−−−=−=−=−+,当1n=时,24410011nn−+=−+=,综上:2441nann=−+,nN,所以2122nnnnanb−==,所以23135212222−=++++nnnT①,①×12得:2341
11352122222+−=++++nnnT②,两式相减得:2341111222221323222222222nnnnnnT++−+=++++−=−,所以2332nnnT+=−故答案为:2332nn+−例8.(2022·全国·高三专题练习)在平面四边形ABCD中,ABD△
的面积是BCD△面积的2倍,又数列na满足12a=,当2n时,恒有()()1122nnnnBDaBAaBC−−=−++,设na的前n项和为nS,则所有正确结论的序号是___________.①na为等比数列;②na为递减数列;③2nna为等差数列;④()15
2210nnSn+=−−【答案】②③④【解析】设AC与BD交于点E,1sin221sin2ABDCBCBDAESEBSAESCEBDCECEB===△△,()22123333BEBAAEBAACBABCBABABC=+=+=+−=+,B,E,D共线,所以存
在实数()0,使得BDBE=,所以()()11122233nnnnBDaBAaBCBABC−−=−++=+,所以11123223nnnnaa−−−=+=,所以()11222nnnnaa−−+=−,1122nnnaa+−=−,所以12a=,24a=−,32
4a=−,na不是等比数列,①错;因为1122nnnaa+−=−,所以11222nnnnaa−−=−,即11222nnnnaa−−−=−,所以2nna是等差数列,③正确;又因为12a=,则112a=,即()1
21322nnann=−−=−,()322nnan=−,所以当2n时,()()()11132232121220nnnnnaannn−−−−=−−−−=−,即1nnaa−,所以na是递减数列,②正确;因为()()()2312121232322nnnS
aaan=+++=+−+−++−,()()()23121212522322nnnSnn+=+−++−+−,所以两式相减得()()()()2312222222322nnnSn+−=+−+−++−−−()()()()211121222
3221052212nnnnn−++−=+−−−=−−−,所以()152210nnSn+=−−,④正确.故答案为:②③④.例9.(2022·云南师大附中高三阶段练习)已知数列na的前n项和为nS,21nnSa=−.(1)求数列
na的通项公式;(2)若数列nb满足2lognnnaba=,求数列nb的前n项和nT.【解析】(1)因为21nnSa=−,所以()*11212,nnSann−−=−N,所以()*11222,
nnnnnaSSaann−−=−=−N,所以()*122,nnaann−=N,当1n=时,11121aSa==−,11a=,所以数列{}na是首项11a=,公比2q=的等比数列,所以12nna-=
;(2)由2lognnnaba=得12211loglog2122nnnnnnanba−−−−===,所以2101211222nnnT−−=++++,231101221222222nnnnnT−−−=+++++,两式相减,得211111122222nnnnT−−=+++−,111111221
12212nnnnn−−−+=−=−−,所以1122nnnT−+=−.例10.(2022·全国·模拟预测(文))若数列na满足221nnnaaa++=,13a=,23243aa=.(1)求na的通
项公式;(2)若3lognnba=,求数列nnab的前n项和nS.【解析】(1)因为数列na满足221nnnaaa++=,13a=,23243aa=,所以0na.所以数列na为等比数列,设其公比为q(0q).所以2232
3113243aaaqaqq===,解得:3q=.所以113nnnaaq−==.即na的通项公式为3nna=.(2)由(1)可知:33l3logognnnban===,所以3nnnabn=,所以1122nnnSababab=+++1213233n
n=+++①3①得:231313233nnnS+=+++②①-②得:()123111313131333nnnSn+−=++++−()1133331133nnnnS+−=−−−所以()13
2134nnSn++−=例11.(2022·全国·模拟预测)已知等差数列na的前n项和为nS,数列nb为等比数列,且111ab==,32312Sb==.(1)求数列na,nb的通项公式;(2)若1nnncab+=,求数列nc的前n项和nT.【解析】(1)设等差数列na的公差
为d,等比数列nb的公比为q,由题意得:13312ad+=,解得:3d=,所以()13132nann=+−=−,由2312b=得:24b=,所以214aqa==,所以14nnb−=(2)()1324nnnncabn+==−,则()2344474324nnTn=++++−①,()23414
44474324nnTn+=++++−②,两式相减得:()23413434343434324nnnTn+−=+++++−−()()111164433241233414nnnnn+++−=+−−=−+−−,所以
()1414nnTn+=+−例12.(2022·全国·高三专题练习)已知数列{na}为等差数列,23a=,1453aa=,数列{nb}的前n项和为nS,且满足231nnSb=−.(1)求{na}和{nb}的通项公式;(2)若nnnc
ab=,数列{nc}的前n项和为nT,且()31nnnTnm−−对nN恒成立,求实数m的取值范围.【解析】(1)解:等差数列{na}中,设公差为d,则211451133313312aadaaadad=+==+=
+()111312122nadaannNadd++===−==数列{nb}中的前n项和为nS,且231nnSb=−①当1n=时,11b=当2n时,11231nnSb−−=−②②-①得:132)(nnbbn−=故数列{
nb}是以1为首项,3为公比的等比数列,所以()13nnbnN−+=.(2)解:数列{nc}中,()1213nnnncabn−==−.则()()01211333233213nnnTnn−−=+++−+−所以()()12131333233213nnnTnn−=+++−
+−故()()()11110221233...321312333nnnnTn−−−=++++−−=−++++()()()1321312213223213nnnnnnn−−−=−+−−=−−−所以()131nnTn=−+∵()1313nnnnmTn−−=−对nN恒成立.当n
为奇数时,()()1min1133131312nnnnmmmm−=−−−−=−=,当n为偶数时,()()22max11313138nnmmm−=−−=−=−综上:实数m的取值范围为()82m−,.【方法技巧与总结】错位相减
法求数列{}na的前n项和(1)适用条件若{}na是公差为()0dd的等差数列,{}nb是公比为()1qq的等比数列,求数列{an·bn}的前n项和nS.(2)基本步骤(3)注意事项①在写出nS与nqS的表达式时,应
特别注意将两式“错位对齐”,以便下一步准确写出nnSqS−;②作差后,应注意减式中所剩各项的符号要变号.等差乘等比数列求和,令()nnqBAnc+=,可以用错位相减法.nnqBAnqBAqBAqBAT)(...)3()2()(32+
+++++++=①1432)(...)3()2()(+++++++++=nnqBAnqBAqBAqBAqT②①②−得:123(1)()()(...)+−=+−+++++nnnqTABqAnBqAqqq.整理得:q
qAqBqqAqBqAnTnn))1(1())1(11(212−−−−−−−+−=+.题型四:分组求和法例13.(2022·广西柳州·模拟预测(理))已知数列{na}满足11a=,()*121Nnnaan+=+.(1)证明{1na+}是等比数列,并求{na
}的通项公式;(2)求数列{1]nan++的前n项和nS.【解析】(1)由题意可得:1120a+=∵()11121212111nnnnnnaaaaaa+++==++=+++所以1na+是首项为2,公比为2的等比数列则12nna+=,即21nna=−因此{na}的通项公
式为21nna=−(2)由(1)知21nna=−,令1nnban=++则2nnbn=+所以()()()121221222nnnSbbbn=+++=++++++.()12222(12)nn=+++++++()()2121122nnn−+=+−()
11222nnn++=+−.综上()11222nnnnS++=+−.例14.(2022·青海·海东市第一中学模拟预测(文))已知正项数列na满足2123232naaanann++++=+,且()()211nnnnabn
n+−=++.(1)求数列na的通项公式;(2)求数列nb的前n项和nS.【解析】(1)解:因为2123232naaanann++++=+,①当2n时,2123123(1)(1)2(1)naaanann−++++−=−+−.②①−②得21n
nan=+,所以21nnan+=.当1n=时,13a=,也满足上式,所以21nnan+=.(2)解:因为(2)(1)1nnannbnn+−=++,则221211111111(1)(1)1nnanbnnnnnnnnnnnnn+=++−=++−=+−=+−−++++,则11111(
3)2311223121nnnnSnnnn+=++++−−+−++−=−++.例15.(2022·上海松江·二模)在等差数列{}na中,已知1210aa+=,34530aaa++=.(1)求数列{}na的通项公式;(2)若数
列{}nnab+是首项为1,公比为3的等比数列,求数列{}nb的前n项和nS.【解析】(1)设等差数列{}na的公差为d,由1210aa+=,34530aaa++=可得112103930adad+=+=,解得
142ad==,∴42(1)22nann=+−=+;(2)∵数列{}nnab+是首项为1,公比为3的等比数列,∴13nnnab−+=,又22nan=+,可得1322nnbn−=−−,所以1(1393)(46
22)nnSn−=++++−++++(13[42]2)113nnnn−=+−−−231322nnn=−−−.【方法技巧与总结】(1)分组转化求和数列求和应从通项入手,若无通项,则先求通项,然后通过对通项变形,转化为等差数列或等比数列或可求前n项和的数列求和.(2)分组转化法求
和的常见类型题型五:裂项相消法例16.(2022·全国·高三专题练习)记nS为数列na的前n项和,已知11,nnSaa=是公差为13的等差数列.(1)求na的通项公式;(2)证明:1
21112naaa+++.【解析】(1)∵11a=,∴111Sa==,∴111Sa=,又∵nnSa是公差为13的等差数列,∴()121133nnSnna+=+−=,∴()23nnnaS+=,∴当
2n时,()1113nnnaS−−+=,∴()()112133nnnnnnanaaSS−−++=−=−,整理得:()()111nnnana−−=+,即111nnanan−+=−,∴31211221nnnnnaaaaaaaaaa−−
−=()1341112212nnnnnn++==−−,显然对于1n=也成立,∴na的通项公式()12nnna+=;(2)()12112,11nannnn==−++∴12111naaa+++1111112121222311nnn=−
+−+−=−++例17.(2022·全国·高三专题练习)记nS为数列na的前n项和,已知11a=,且13nnSa+=−.(1)求数列na的通项公式;(2)已知数列nc满足________,记nT为数列
nc的前n项和,证明:2nT.从①211(1)(2)nnnncaaa+++−−=②221lognnnaca++=两个条件中任选一个,补充在第(2)问中的横线上并作答.【解析】(1)13nnSa+=−①,当1n=时,123aa=−,24a=;当2n时,13nnSa−=−②①-②得,
即12nnaa+=又2142aa=,∴数列na是从第2项起的等比数列,即当2n时,2222nnnaa−==.1,1,2,2.nnnan==.(2)若选择①:()()()()()()22111
11122211212212121222121nnnnnnnnnnnnacaa++++++++====−−−−−−−−−,2231111111121212212121212121nnnnT++=−+−++−=
−−−−−−−.若选择②122nnnc++=,则23134122222nnnnnT+++=++++③,34121341222222nnnnnT++++=++++④,③-④得3412121311123112124
22224422nnnnnnnT++−+++=++++−=+−−,14222nnnT++=−.例18.(2022·全国·高三专题练习(理))已知正项数列{na}中,11a=,nS是其前n项和,且满足()211nnSSS+=
+(1)求数列{na}的通项公式:(2)已知数列{nb}满足()1111nnnnnabaa+++=−,设数列{nb}的前n项和为nT,求nT的最小值.【解析】(1)正项数列{na},11a=,满足()211nnSSS+=+,所以11nnSS+−=,所以数列{nS}是以1为首项1为公差的等
差数列,所以1(1)1nSnn=+−=,所以2nSn=,当2n时,221(1)21(N*)nnnaSSnnnn−=−=−−=−,当1n=时也成立,所以21(N*)nann=−.(2)因为()1111nnnnnabaa
+++=−()()()()1112111212122121nnnnnnn++−=−=+−+−+所以1111111111(1)()(1)()1(1)23352121221nnnTnnn++=+−+++−+=+−
−++,所以当n为奇数时,11112212nTn==++()>;当n为偶数时,111221nTn==−+(),由{nT}递增,得225nTT=,所以nT的最小值为25.例19.(2022·浙江·模拟预测)已知数列na的首项为正数,其前
n项和nS满足82343nnnnSaSa=−−.(1)求实数的值,使得2nS+是等比数列;(2)设13nnnnbSS+=,求数列2nb的前n项和.【解析】(1)当1n=时,111823aaa=−,11Sa=,解得22118Sa==;当2
n时,把1nnnaSS−=−代入题设条件得:22198nnSS−=+,即()221191nnSS−+=+,很显然21nS+是首项为8+1=9,公比为9的等比数列,∴1=;(2)由(1)知21nS+是首项为21
190S+=,公比9q=的等比数列,所以291nnS=−,()()()()()()1211191919111188919919199111nnnnnnnnnnb++++−−−===−−−−−−−.故数列2nb
的前n项和为:2221122334112111111111111891919191919191918891nnnnbbb+++++=−+−+−++−=−−−−−−−−−−.例20.(2022·湖南·一模)已知等差数列{}na中,前n项和
为nS,11a=,{}nb为等比数列且各项均为正数,11b=,且满足227bS+=,3322bS+=.(1)求na与nb;(2)设32nnnbc−=,1(2)(1)(1)nnnnnnnaadaac++=−+,求nd的前2n项和2nT.【解析】(1)由题意,设等差数列{}na的公差为d,等比数列{
}nb的公比为(0)qq,则212122Saaadd=+=+=+,312313333Saaaadd=++=+=+,21bbqq==,2231bbqq==,227bS+=,3322bS+=,227332
2qdqd++=++=,即25319qdqd+=+=,解得16qd=−=(舍去),或41qd==,1(1)1nann=+−=,*nN,11144nnnb−−==,*nN.(2)由(1),可得1133
4222nnnnnnbc−+−−===,则111(2)2(1)11(1)(1)(1)[](1)(1)22(1)2nnnnnnnnnnnnaanndaacnnnn++++++=−=−=−++++,21232nnTdddd=++++12233422111
11111112222232324222(21)2nnnn+=−−++−−++++2111(21)22nn+=−+.例21.(2022·全国·高三专题练习)已知数列na前n项和为nS,且()21nSnn=+,记221(1)nnnnnab
aa+=−+.(1)求数列na的通项公式;(2)设数列nb的前n项和为nT,求2021T.【解析】(1)()112nSnn=+,当1n=时,111212S==;当2n,nN时,()1112nSnn−=−,()()1111122
nnnaSSnnnnn−=−=+−−=.当1n=时也符合,()nannN=.(2)()()()()()()221212111111111nnnnnnnnnnanbaannnnnn++++=−=−=−=−++++
+202111111111...12233420212022T=−+++−++−+111111112023=1...1223342021202220222022−−++−−+−−=−−=−
.例22.(2022·河南·洛宁县第一高级中学一模(文))已知数列na是公差不为零的等差数列,2414aa+=,且1a,2a,6a成等比数列.(1)求na的通项公式;(2)设11nnnbaa+=,求数列nb的前n项和nS.【解析】(1)等差数列na中,324214aaa=
+=,解得37a=,因1a,2a,6a成等比数列,即2216aaa=,设na的公差为d,于是得()()()277273ddd−=−+,整理得230dd−=,而0d,解得3d=,所以()3332naandn=+−=−.(2)由(1)知,()()1
111()323133231nbnnnn==−−+−+,所以111111[(1)()()]34473231nSnn=−+−++−−+11(1)33131nnn=−=++.例23.(2022·山西大同·高三阶段
练习)已知数列na的前n项和nS满足()22NnnSan++=.(1)证明:数列2nS+是等比数列;(2)设数列()()1211nnnaa+−−的前n项和为nT,求证:213nT.【解析】(1)证明:当1n=时,1122Sa+
=∴112Sa==当2n时,()122nnnSSS−+=−122nnSS−=+,()1222nnSS−+=+∴1222nnSS−+=+∴数列2nS+是以2为公比,首项124S+=的等比数列(2)由(1)知1242nnS−+=,122nnS+=−,代入22nnSa+=
得2nna=()()1121121212121nnnnn++=−−−−−∴2231111111212121212121nnnT+=−+−++−−−−−−−111121n+=−−
由1n,124n+,1213n+−111213n+−,所以111213n+−−−∴1121213n+−−综上所述213nT例24.(2022·江西九江·三模(理))已知数列na的前n项和为nS,且满足12a=,1436nnnaaS++
=+.(1)求na;(2)求数列()21nnnna++的前n项和.【解析】(1)当1n=时,211436aaS+=+,∵12a=,∴24a=.当2n时,由1436nnnaaS++=+,得11436nnnaaS−−+=+
,两式相减得()1143nnnnnaaaaa+−−+−=即114nnaa+−=∴数列21na−,2na均为公比为4的等比数列∴12121242nnna−−−==,122442nnna−==∴2nna=(2)∵()()()122112112212nnnnnnnnannn
n+++==−+++∴数列()21nnnna++的前n项和()12231111111212222232212nnnTnn+=−+−++−+()()111112
1121212nnnn+=−=−++例25.(2022·广东·大埔县虎山中学高三阶段练习)已知各项均不相等的等差数列na的前4项和为10,且124,,aaa是等比数列nb的前3项.(1)求,nna
b;(2)设22121nnnnncbaa++=+,求nc的前n项和nS.【解析】(1)设等差数列na的公差为d,0d,则12214434102adaaa+==,得()()121112353adadaad+=+=
+,得121235addad+==,因为0d,所以11235adda+==,解得11ad==,所以1(1)naandn=+−=,所以111ba==,22112ba==+=,所以等比数列nb的公比212
aqa==,所以12nnb−=.(2)122212(1)nnncnn−+=++122112(1)nnn−=+−+,所以2122222211111112221223(1)nnSnn−=+++++−+−++−+2121112(1)nn−=+−−+2
12(1)nn=−+.例26.(2022·全国·高三专题练习)等比数列na中,首项11a=,前n项和为nS,且满足()1344aaS+=.(1)求数列na的通项公式;(2)若31(1)log+
=+nnbna,求数列242nnb+的前n项和nT.【解析】(1)设数列na公比为q,由11a=,()1344aaS+=,可得32330qqq−+−=,化简得()()2130qq+−=,即3q=,所以13−=nna.(2)由
(1)得3(1)log3(1)nnbnnn=+=+,所以222224242112(1)(1)nnnbnnnn++==−++所以22222111112122223(1)nTnn=−+−++−+()
()22222211111221222311nnn=−+−++−=−++..例27.(2022·全国·高三专题练习)已知等差数列na的前n项和为nS,且11a=,5212SS=+;数列nb的前n项和nT,且11b=,数列nb的11nnbT+=+,
()*nN.(1)求数列na、nb的通项公式;(2)若数列nc满足:()()112141nnnnnnnaacaab−++=−+,当2n时,求证:12212nccc+++.【解析】(1)解
:因为11a=,由5212SS=+,得34512aaa++=,所以4312a=,即44a=,设等差数列na的公差为d,所以41141aad−==−,所以()()11111naandnn=+−=+−=.由11nnbT+=+,()*nN,得11nnbT−=+,()
2n,两式相减得()11nnnnnbbTTb+−−=−=,即()122nnbbn+=,又2111112bTbb=+=+=,所以数列nb是以1为首项、2为公比的等比数列,则11122nnnbb−−==;(2)由(1)知:()()()()()(
)1111221114112nnnnnnnnnaanncaabnn−−+++++=−=−++,()()11111212nnnnn−+=−++,∴21232122334111111122222323242nnTcccc=++++=+−+++−
()()22121111112221222122nnnnnn++−+=−++.例28.(2022·广东惠州·高三阶段练习)记nS是公差不为零的等差数列na的前n项和,
若36S=,3a是1a和9a的等比中项.(1)求数列na的通项公式;(2)记121nnnnbaaa++=,求数列nb的前20项和.【解析】(1)由题意知2319aaa=,设等差数列na
的公差为d,则()()211182aadad+=+,因为0d,解得1ad=又31336Sad=+=,可得11ad==,所以数列na是以1为首项和公差为1的等差数列,所以()11naandn=+−=,*Nn(2)由(1)可知()()()()()1
111122112nbnnnnnnn==−+++++,设数列nb的前n和为nT,则()()()1111111212232334112nTnnnn=−+−++−+
++()()1112212nn=−++,所以20111115222122462T=−=所以数列nb的前20和为115462例29.(2022·河北衡水·高三阶段练习)已知数列na
的前n项和为nS,且满足244nnSa=−,数列nb满足11ba=,1nnnbba+−=,*nN.(1)求数列na,nb的通项公式;(2)设11()(1)nnnnbcanan++=+++,且数列nc的前n项和为nT,若nkT,恒成立,求常数k的最小
值.【解析】(1)由244nnSa=−,得当1n=时,12a=,当2n…时,11244nnSa−−=−,两式相减得1244nnnaaa−=−,12nnaa−=,数列na是首项为2,公比为2的等比数列,2nna=.由11ba=,1
2nnnbb+−=,*nN,12a=,得212bb−=,2322bb−=,…,112nnnbb−−−=,累加得2311222...2nnbb−−=++++12(1212)n−−=−22n=−,2nnb=,*nN.
(2)由(1)得()()()()111121111221221nnnnnnnnnbcanannnnn+++++===−+++++++++,123...nnTcccc=++++2231111111...21222223221nnnn+=−+−++−+++++
++11311213nn+=++−,13k…,即常数k的最小值为13.例30.(2022·全国·高三专题练习)已知等比数列na公比为正数,其前n项和为nS,且4244,30aaS==.数列nb满足:*1115,23,2nnnnbababnnN++==++.(1)求数列,
nnab的通项公式:(2)求证:()()3112..212233411nnbbbbbnnnn−+++++−+.【解析】(1)()413111112,302,221aqaqaqaqq−====−2nna=当2n时,()()()1111222211nnnn
nnnnabababababab−−−−−−−+−++−()()21322323nn=−++−++++,()()()111112312nnnnababn−+−−=+−222nnabnn=++,又2nn
a=()222,22nnnnbn++=,经检验152b=符合上式,2222,2nnnnnnab++==(2)()()()221222211212nnnbnnnnnnnnnnn++++++==+++.()1111122212nnnnn++=+
−+()()3112..12233411nnbbbbbnnnn−+++++−+()21232311111111111221222222322212nnnnn++=+−++−+++−+.(
)211111221121121212nnn+−=+−+−()1111212.212nnn++=−−+另解:()()()21122122112212nnnnbnnnnnnnnnn++++
++==−+++()()311212233411nnbbbbbnnnn−+++++−+()2231233412212222232212nnnnnn+++=−+−++−+()1221212nn
n++=−+.得证例31.(2022·广东佛山·二模)已知数列{na}的前n项和为nS,且满足()()*1311,N,5nnnSnSnnna+−+=+=(1)求1a、2a的值及数列{na}的通项公式na:(2)设1nnnbaa+=,求数列{
nb}的前n项和nT【解析】(1)因()()*1311,N,5nnnSnSnnna+−+=+=,取1n=和2n=得:12112312()222()3()6aaaaaaaa+−=++−+=,即2112
24aaaa−=+=,解得121,3aa==,由()()111nnnSnSnn+−+=+得:111nnSSnn+−=+,数列{}nSn是首项为1111Sa==,公差1d=的等差数列,则nSnn=,即2nSn=,当2n时,221(1)21nnnaSSnnn−=−=−−=−,而11a=
满足上式,因此,21nan=−,所以121,3aa==,数列{na}的通项公式21nan=−.(2)由(1)知,当2n时,()121112116nnnnnnnnnnnnaaaaaaaaaaaa++−+++−+−=−=6nb=,因此,12
111()6nnnnnnnbaaaaaa++−+=−,1212312211()(15)66nknnnnnnkbaaaaaaaaa++++==−=−,则21121221151(15)(21)(21)(23)3(461)6623nnknnnkTbbaaaaann
nnnn++==+=−+=−++−+=+−,13b=满足上式,所以21(461)3nTnnn=+−.例32.(2022·全国·高三专题练习)已知正项数列na的前n项和为nS,且满足11a=,23a=,213
2nnnaaa++=−,数列nc满足()22221232341ncccncn+++++=.(1)求出na,nc的通项公式;(2)设数列()()1221log1nncna+++
的前n项和为nT,求证:516nT.【解析】(1)由2132nnnaaa++=−,得()2112nnnnaaaa+++−=−.又212aa−=,则数列1nnaa+−是首项为2,公比为2的等比数列,∴11222nnnn
aa−+−==,∴221322,2aaaa−=−=,3432aa-=,…,112nnnaa−−−=,累加得211222nnaa−−=+++,∴211212222112nnnna−−=++++==−−.数列nc满足()22221232341ncc
cncn+++++=,①当1n=时,114c=;当2n时,222212312341ncccncn−++++=−,②由①-②可得()211ncn=+,当1n=时,也符合上式,故数列nc的通项公式为()211ncn=+.(2)由(1)可得()()()
()2222222111114222log1nnnnnnnna++==−++++,则()22222221111111114324352nTnn=−+−+−++−+()()222111114212nn=+−−++
()()2215115441612nn=−−++,故516nT成立.例33.(2022·天津南开·三模)已知数列na是公比1q的等比数列,前三项和为13,且1a,22a+,3a恰好分别是等差数列nb的第一项,第三项,第五
项.(1)求na和nb的通项公式;(2)已知*kN,数列nc满足21,21,2nnnnnnkbbcabnk+=−==,求数列nc的前2n项和2nS;(3)设()()2(810)12121nnnnnadaa+−−=++,求数列nd的前n项和nT.【解
析】(1)(1)解:()()()21123122131113131223124aqqaaaaaaaqaqq++=++==+=+=−+=或1913aq==,又1qQ,则113aq==,∴13−
=nna(n*N).设等差数列nb的公差为d,由题意得,111ba==,322325ba=+=+=,即11111252bbbdd==+==,所以21nbn=−(n*N).(2)(2)解:21nk=−时,21212111111(4
3)(41)44341nkkkccbbkkkk−−+====−−+−+,∴13521nScccc−=++++奇111111111111415459491344341nn=−
+−+−++−−+111111111415599134341nn=−+−+−++−−+11144141nnn=−=++.2nk=时,21222(41)3knkkkccabk−===−∴2462nSc
ccc=++++偶151323373113(41)3nSn−=++++−偶,①53212193373(45)3(41)3nnSnn−+=+++−+−偶,②由①−②可得,1321215833434343(41
)3nnSn−+−=++++−−偶21321434399(41)319nnn−+=+−−−−∴21916(83)3nnS+−+=偶∴2126(83)34911nnnnSSSn+−=+=+++偶奇(*nN).(3)(3)由(
1)知13−=nna,则()()()()111112(810)1(810)3111121212231231231231nnnnnnnnnnannndaa−−+−++−−−−−+===−++++++∴0213241021131242
23123122312312231231nT=−+−+−++++++111112231231nnnn−+−+++−++01110112231231231231nnnn++=+−
−++++1111142231231nnnn++=−+++故1111142231231nnnnnT++=−+++(*nN).【方法技巧与总结】裂裂项相消法求
和(1)基本步骤(2)裂项原则一般是前边裂几项,后边就裂几项,直到发现被消去项的规律为止.(3)消项规律消项后前边剩几项,后边就剩几项,前边剩第几项,后边就剩倒数第几项.题型六:倒序相加法例34.(2022·河北·高三阶段练习)德国大数
学家高斯年少成名,被誉为数学届的王子,19岁的高斯得到了一个数学史上非常重要的结论,就是《正十七边形尺规作图之理论与方法》.在其年幼时,对123100++++的求和运算中,提出了倒序相加法的原理,该原理基于所给数据前后对应项的和呈现一定的规律生成,因此,此方法也称之为高斯算法,现有函数2()2
2xxfx=+,设数列na满足()121(0)(1)Nnnafffffnnnn−=+++++,若12nnnba+=,则nb的前n项和nS=_________.【答案】12nn+【解析】由
2()22xxfx=+得,1122()(1)222222212222222222xxxxxxxxxxfxfx−−+−=+=+=+=++++++,由()121(0)(1)Nnnafffffnnnn−=+++++,得121(1)(0)nnafff
ffnnn−=+++++,故121,2nnnana+=+=,故12(1)2nnnnban+==+,所以123223242...(1)2nnSn=+++++,则312422232
422(1)2nnnSnn+=++++++,两式相减得:2312222...2(1)2nnnSn+−=++++−+112(12)221122()nnnnn++−+=−−=+−故12nnSn+=,故答案为:12nn+例35.(202
2·黑龙江齐齐哈尔·三模(文))已知数列na的前n项和为nS,且1211121nnSSSn+++=+,设函数()1cosπ2fxx=+,则32021122022202220222022aaaaffff+
+++=______.【答案】20212【解析】∵1211121nnSSSn+++=+①,∴当2n时,()12121111nnSSSn−−+++=②,①-②得()121nS
nn=+,∴()()122nnnSn+=;当1n=时,111S=,∴11S=,此时()12nnnS+=仍然成立,∴()()*12nnnSn+=N.∴当n=1时,111aS==;当2n时,()()11122nnnnnnnaSSn−+−=
−=−=,当n=1时,上式也成立,故nan=()*nN.由于()()()111cosπcosππ122fxfxxx+−=++−+=,设32021122022202220222022aaaaSffff=++
++12320212022202220222022ffff=++++则120212202020211220222022202
2202220222022Sffffff=++++++2021=,∴20212S=.故答案为:20212.例36.(2022·全国·高三专题练习(文
))已知数列na,nb满足1118a=,11216nnnnaaaa++−=,116nnba=−.(1)证明nb为等比数列,并求nb的通项公式;(2)求11223377abababab++++L.【解析】(1)由11216nnnnaaaa++−=可得1121
6nnaa+=−,于是11116216nnaa+−=−,即12nnbb+=,而111162ba=−=,所以nb是首项为2,公比为2的等比数列.所以1222nnnb−==.(2)由(1)知1216nna=+,所以2216nnnnab=+.因
为8488844222112162162122kkkkkkkkkkkabab−−−−−−−+=+=+=++++,所以()112233772abababab++++L()()()1177226677117abababababab=++
++++=L,因此1122337772abababab++++=L.例37.(2022·全国·高三专题练习)已知函数()21122fxxx=+,数列na的前n项和为nS,点()()*,NnnSn均在函数()fx的图象上,函数()442xxgx=+.(1)求数列n
a的通项公式;(2)求()()1gxgx+−的值;(3)令()*2021nnabgn=N,求数列nb的前2020项和2020T.【解析】(1)因为点()()*,NnnSn均在函数()fx的图象上,所以21122nSn
n=+,当2n时,()()2211111112222nnnaSSnnnnn−=−=+−−−−=,当1n=时,111aS==,适合上式,所以nan=.(2)因为()442xxgx=+,所以()114214242xxxgx−−−==++,
所以()()42114242xxxgxgx+−=+=++.(3)由(1)知nan=,可得20212021nnanbgg==,所以2020122020122020202120212021Tbbbggg=+++=+++
,①又因为2020202020191202020191202120212021Tbbbggg=+++=+++,②因为()()11gxgx+−=,所以①+②,得202022020T=,所以20201010T=.例38.(2022·全
国·高三专题练习)已知函数()331xxfx=+,()xR,正项等比数列na满足501a=,则()()()1299lnlnlnfafafa+++值是多少?.【解析】因为3()31xxfx=+,所以33331)331)()()1(3131(31)(31)(xxxxxxxxxxfxfx−−−
−−++++−=+==++++.因为数列na是等比数列,所以21992984951501=====aaaaaaa,即199298491055lnlnlnlnlnln02lnaaaaaaa+=+===+=.199298490515))))1))2(ln(ln(ln(ln(l
n(ln(ln)ffafafafafafaa==++==+=设9912399(ln)(ln)(ln)(ln)=++++Sfafafafa①,又99999897(ln)(ln)(ln)=++Sfafafa+…+1(ln)fa②,①+②,得99299=S,所以99992
=S.例39.(2022·全国·高三专题练习)已知函数()fx对任意的xR,都有()()11fxfx+−=,数列na满足()120nafffnn=+++…()11nffn−+.求数列na的通项公式.【解析】
因为()()11fxfx+−=,111nffnn−+=.故()120nafffnn=+++…()11nffn−++.①()121nnnafffnn−−=+++
…()01fnf++.②①+②,得21nan=+,12nna+=.所以数列na的通项公式为12nna+=.例40.(2022·全国·高三专题练习)已知函数()21122fxxx=+,数列na的前n项和
为nS,点()()*,nnSnN均在函数()fx的图象上.(1)求数列na的通项公式;(2)若函数()442xxgx=+,令()*2021nnabgn=N,求数列nb的前2020项和2020T.【解析】(1)∵点(),nnS均在函数()fx的图象上
,∴21122nSnn=+.当2n时,1nnnaSSn−=−=;当1n=时,111aS==,适合上式,∴nan=.(2)∵()442xxgx=+,∴()()11gxgx+−=.又由(1)知nan=,∴2021nnbg=.∴20201220201220
20202120212021Tbbbggg=+++=+++,①又2020202020191202020191202120212021Tbbbggg=+++=+++,②①+②,202012
02022020202020212021Tgg=+=,∴20201010T=.【方法技巧与总结】将一个数列倒过来排列,当它与原数列相加时,若有规律可循,并且容易求和,则这样的数列求和时可用倒序相加法(等差数列前n项和公式的推导即用此方法).题型
七:并项求和例41.(2022·全国·高三专题练习)已知na的通项公式为()1nnan=−,求na的前n项和nS.【解析】解:当n为偶数时,12345678=−+−+−+−+−+nSn,122==nn;当n为奇数时,()()
()()123421=−++−+++−−+−−nSnnn,1122−+=−=−nnn.综上:()()122nnnSnn+−=为奇数为偶数.例42.(2022·福建·厦门一中模拟预测)已知数列{}na的前n项和nS,11a=,0na,141n
nnaaS+=−.(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.【解析】(1)解:当1n=时,12141aaa=−,解得23a=,由题知141nnnaaS+=−①,12141n
nnaaS+++=−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)432211n
annn−=+−=−=−−,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(4
1)nnbbnnnnn−=−−−+−+=−+21234212(341)()()()4[37(41)]44(21)2nnnnnTbbbbbbnnn−+−=++++++=+++−==+.例43.(2022·河北·沧县中学模拟预测)已知数列
na为等差数列,nS为其前n项和,若34102252,33+==aaS.(1)求数列na的通项公式;(2)若()22π1cos3nnnba=+,求数列nb的前18项和18T.【解析】(1)设等差数列
na的公差为d.则34102252,33aaS+==112383109251023adad+=+=,解得12313ad=−=.故数列na的通项公式为21(1)1333=−+−=−nnan.(2
)由(1)知,13nna=−,所以()2222π2π12π1cos11coscos33393nnnnnnban=+=−+=.因为当Nk时,2223231324π12π5cos2πcos2πcos2π3
33318kkkbbbkkkkkkk−−++=−−+−−+=−,()()()18123345161718Tbbbbbbbbb=+++++++++.555555123456181818181818=−+−+−+−+−+−558(1
23456)6183=+++++−=所以数列nb的前18项和为583.例44.(2022·全国·高三专题练习)已知数列na的前n项和为nS,且满足21322nSnn=+.(1)求na的通项公式;(2)在ka和()*1Nkak+中插入k个相同
的数()11kk+−,构成一个新数列1:nba,1,2a,2−,2−,3a,3,3,3,4a,L,求nb的前21项和21T.【解析】(1)解:因为21322nSnn=+,当1n=时,112aS==,当2n时,
()()22113131112222nnnaSSnnnnn−=−=+−−+−=+,12a=也满足1nan=+,所以,对任意的Nn,1nan=+.(2)解;在ka和()*1Nkak+中插入k个相同的数()11kk+−,构成一个新数列1a,1,2a,2−,2
−,3a,3,3,3,4a,L,1ka+,其项数为()()()()112112122kkkkkkk++++++++=++=,因为67212=,即当5k=时,()()12212kk++=,因此,()()2112662712233445515422Taaa+=++++−+−
+=+=.例45.(2022·河南·汝州市第一高级中学模拟预测(理))在数列na中,15a=,且()*121nnaanN+=−.(1)证明:1na−为等比数列,并求na的通项公式;(2)令
(1)nnnba=−,求数列nb的前n项和nS.【解析】(1)解:因为121nnaa+=−,所以()1121nnaa+−=−,又114a−=,所以1121nnaa+−=−,所以1na−是以4为首项,2为公比的等比数列.故1142nna−−=,即121nna+=+
.(2)解:由(1)得()1(1)21nnnb+=−+,则()1*1*21,2,21,21,nnnnkkNbnkkN+++==−+=−,①当*2,nkk=N时,()()()()()23412
121212121nnnS+=−−++−+++−−++()2345124422222222221;3nnnn+=−+−++−+=+++=−②当*21,nkk=−N时,()()21211427212133nnnnnnSSb++++++=−=−−+=−,综上所述,()*2*421,2,327
,21,3nnnnkkNSnkkN+−==+−=−例46.(2022·全国·高三专题练习)已知数列na满足15a=,214321nnaann+=−++.(1)证明:数列2nan−为等比数列.(2)求数列()1nna
−的前n项和nS.【解析】(1)证明:因为2114a−=,214321nnaann+=−++,所以()()22221222143211444nnnnnnanannnanananan+−+−++−+−===−−−,所以数
列2nan−是首项为4,公比为4的等比数列;(2)解:由(1)可得2144nnan−−=,即24nnan=+,则()()()()23412341111nnnSaaaaa=−+−+−+−++−()()()()()()2223242142434441
4nnn=−−+++−−++++−+.当n为偶数时,()()()122111114142134nnnnnnnaannn−−−−−+−=+−−−=−+,则35133473411342134nnSn−=+++++++−+()351371121344
44nn−=++++−+++++()()()132141414423211652nnnnnn++−−+−=+=+−,当n为奇数时,则()()()22114414452nnnnnnnnSSnn−−−=+−+=+−−()114452nnn+++=−−,综上所述,(
)()11144,52144,52nnnnnnSnnn++++−−=+−+为奇数为偶数.【方法技巧与总结】两两并项或者四四并项题型八:先放缩后裂项求和例47.(2022·天津市宝坻区第一中学二模)已知na为等差数列,前n项
和为(),nnSnNb是首项为2的等比数列,且公比大于0,2334111412,2,11bbbaaSb+==−=.(1)na和nb的通项公式;(2)求数列2nnab的前8项和8T;(3)证明:()212591niiibb=
−.【解析】(1)解:设等差数列na的公差为d,等比数列nb的公比为q.由已知2312bb+=,得()2112bqq+=,而12b=,所以260qq+−=.又因为0q,解得2q=.所以2nnb=.由3412baa=−,可得138da−=①.由11411
Sb=,得1516ad+=②,联立①②,解得11,3==ad,由此可得32nan=−.所以,na的通项公式为32,=−nnanb的通项公式为2nnb=.(2)解:设数列2nnab的前n项和为nT,由262nan=−,得2(62)2=−nnna
bn,所以2342102162(62)2nnTn=++++−,2341242102162(68)2(62)2nnnTnn+=++++−+−,上述两式相减,得23142626262(62)2nnnTn+−=+++
+−−()1212124(62)2(34)21612nnnnn++−=−−−=−−−−.得2(34)216nnTn+=−+.所以,数列2nnab的前n项和为2(34)216,nnTn+=−+
当8n=时,820496=T.(3)解:由(1)得2nnb=,所以:当1n=时,()12212252(21)91==−−bb,不等式成立;当2n=时,()222244(41)91==−−bb,所以()
()12221242225299911+=+=−−bbbb,不等式成立;当3n时,()()()()()11122222222211111211122nnnnnnnnnn−−−==−−−−−−−−,所以,()()()()()12322222112311
111==++++−−−−−niniinbbbbbbbbbb23341411111129212121212121−=++−+−+−−−−−−−nn2221125125921219219=+−=−−−−nn,所以()2
12591niiibb=−,得证.例48.(2022·浙江·效实中学模拟预测)设各项均为正数的数列na的前n项和为nS,满足()()222*330,nnSnnSnnnN−+−−+=.(1)求1a的值:(2)求数列
na的通项公式:(3)证明:对一切正整数n,有112211122211442221nnaaaaaann++++−+++++.【解析】(1)令1n=,()()1121133101−+−−+=SS,则13a=−舍去,所以12a=.(2)(
)()()()2222330,30nnnnSnnSnnSSnn−+−−+=+−−=,因为数列na各项均为正数,3−nS舍去,2=+nSnn,当2n时,()()21111,2−−===−+−−nnnnSnnaSSn,12,12.2,2−===−=nn
nnnaanSSnn(3)令()111222221211nbnnnnnnn===+++++()()()()211211112111nnnnnnn=−+++−−++−()()()1121124112211n
nnnnnn+−−==−−+−+,所以1212111111432411nnSbbbbnn=++++−+−++−−+12211222111.4424411nnnn+=++−−=−+++
例49.(2022·广东汕头·一模)已知数列na的前n项和为nS,()*322nnaSnnN=+.(1)证明:数列1na+为等比数列,并求数列na的前n项和为nS;(2)设()31log1nnba+=+,证明:2221
21111nbbb+++.【解析】(1)当1n=时,11322aS=+,即12a=由322nnaSn=+,则()1132212nnaSnn−−=+−两式相减可得13223nnnaaa−=+−,即132nnaa−=+所以()1131nnaa−+=
+,即1131nnaa−+=+数列1na+为等比数列则()112133nnna−+=+=,所以31nna=−则()()1231333333132nnnnnnS+−−=+++−==−−L(2)()1313log1log31nnnban++
=+==+()()2211111111nbnnnnn==−+++所以2221211111111111122311nbbbnnn+++−+−++−=−++L例50.(2022·浙江绍
兴·模拟预测)已知等差数列na的首项为11a=,且1533+=+aaa,数列nb满足1122(21)31,2nnnnabababn−++++=N.(1)求na和nb;(2)设2=−nnncba,记12111nnTccc=+++,证明:当n
N时,216473nnnTba++−.【解析】(1)因为na是等差数列,设其公差为d.因为1533+=+aaa,所以111342addaa+=+++.因为11a=,所以等差数列na的公差1d=,所以1
(1)naandn=+−=.因为1122(21)312nnnnababab−++++=,所以112ab=,所以12b=.当2n时,11(21)31(23)312322−−−+−+=−=nnnnnnnabn,结合
nan=可知123nnb−=.经检验:12b=也适合上式.所以123nnb−=.(2)由(1)可知:221323(1)+−=−+nnnban.所以要证明原不等式成立,只需证明:27314223(1)−−+nnTn成立.易得:21223−=−=−nnnncban,所以当1n=时
,左边111==c,右边1=,左边=右边.当2n时,2231(1)(21)022+−+=−−nnnn,此时2123(1)343(21)2−−+−+nnnn.所以()2121221123(1)232323(1)−−−+=
=−−−+nnnnnncnnn2112223(1)1143(21)2323(1)−−−+=−−+−−+nnnnnnnn12231122323(1)−−−−+nnnn所以1222121113
117311223223(1)4223(1)++++−−−−+−+nnncccnn于是,当nN时,27314223(1)−−+nnTn成立.综上所述:当nN时,216473++−nnnTba.例51.(2022·天津·一模)已知数列n
a是等差数列,其前n项和为nA,715a=,763A=;数列nb的前n项和为nB,()*233nnBbn=−N.(1)求数列na,nb的通项公式;(2)求数列1nA的前n项和nS;(3)求证:12nkkkaB=.【解析】(1)数列na是
等差数列,设公差为d,7171615767632aadAad=+==+=,化简得1161539adad+=+=,解得13a=,2d=,∴21nan=+,*nΝ.由已知233nnBb=−,当1n=时,1112
332Bbb=−=,解得13b=,当2n时,11233nnBb−−=−,∴()()11122333333nnnnnnBBbbbb−−−−=−−−=−,*nΝ,即13nnbb−=,∴数列nb构成首项为3,公比为3的等比数列,∴3nnb=,*
nΝ.(2)由(1)可得()()()1321222nnnaannAnn+++===+,*nΝ,∴()11111222nAnnnn==−++,∴()()()()1111111324352112nSnnnnnn=++++
++−−++11111111111112324352112nnnnnn=−+−+−++−+−+−−−++()()111131113231221242124212nnnnnnn+=+−−=−+=−++++
++(3)由(1)可得()()313331132nnnB−−==−,*nΝ,则()212213313312nnnnannB++==−−,方法一:∵111131331233123nnnnn−−−−−=−=+−≥,∴1221221213313
233nnnnnannnB−+++==−≤,令213521213333nnnnnT−−+=++++,231135212133333nnnnnT+−+=++++,两式相减可得2312111211233
333nnnnT++=++++−11111111211121424931213139333313nnnnnnnn−++++−+++=+−=+−−=−−,∴223nnnT+=−
,∴2312214161212223313131313nknnkkannB=+++++=++++−−−−−≤方法二:∵2n时,()0101223112122122121nnnnnnnnnnCCCCCCn−=+−=++
+−++−+≥,根据“若0ab,0m,则bbmaam++”,可得()2121313nnnn++−≤,∴()23212122141612123233313231313133nknnkknanB=+++++=+++++++−−−−
≤,令()23212324333nnT+=+++,()3412112324233333nnnnT++=++++,两式相减可得()3412122212121333333nnnT++=++++−()()32111211212127172533
1339339313nnnnnnnn−+++−+++=+−=−−=−−,∴725623nnT+=−∴76T,∴23122141612123728233131313132633nknkkanB=++++=+++++=
−−−−方法三:令2131nnnc+=−,下一步用分析法证明“112nncc+”要证112nncc+,即证()()()()12331123121nnnn++−−+,即证()()()()
146312131nnnn++−+−,即证()25233nnn−−−,当*nN,显然成立,∴112nncc+,∴()1121223521233131332831322222nnknnkkancccB−=+=+++=++++++
−≤11232123121323212nn−==−−例52.(2022·全国·高三专题练习)求证:11114313213217n−++++++
.【解析】111132132nn−−+,21111131321321321n−++++++++2121111111114732321283232nn−−=++++++++2121111111111128
32228322nn−−=+++=+++211111111114748442411283283848471122n−−=++==−−【方法技巧与总结】先放缩后裂项,放缩的目的是为了“求和”,这也是凑配放缩形式的目标.题型九:分段数列
求和例53.(2022·全国·高三专题练习)设数列na的前n项和为nS,且满足22nnSa=−.(1)求数列na的通项公式;(2)若*222,21,N22,2,Nloglognnnnankkbnkkaa+=+==+,求数列nb
的前15项的和.【解析】(1)由22nnSa=−得,当n=1时,1122Sa=−,解得12a=.当n≥2时,1122nnSa−−=−,从而122nnnaaa−=−,即12nnaa−=,因此数列na是等比数列,其首项
和公比都等于2,所以2nna=.(2)当n为奇数时,122nnb−=,当n为偶数时,222nbnnnn==+−++,所以数列nb的前15项和为()()121513152414bbbbbbbbb++=+++++()()()()27122242641614=+++++−
+−++−81216212−=+−−2592=−.例54.(2022·山东师范大学附中模拟预测)已知nS是数列na的前n项和,且111,21nnaaan+=+=+.(1)求数列na的通项公式;(2)记2,,nannnban=为奇
数为偶数,求数列nb的前2n项和nT.【解析】(1)121++=+nnaan变形为()()11nnanan+−+=−−,因为110a−=,所以()()()111?··10nnanana+−+=−−==−−=,故nan=;(2)当n为奇数时,2nnb=,当n为偶数时,nbn=,
则3521222242622nnTn−=++++++++()352124622222nn−+++++++++()2121222222221433nnnnnn+++−=+=−++−例55.(2022·湖南·长郡中学模拟预测)已知数列na满足12a=,
11,22,nnnanaan+−=+为奇数为偶数.(1)记21nnba−=,证明:数列nb为等比数列,并求出数列nb的通项公式;(2)求数列na的前2n项和2nS.【解析】(1)因为12a=,当n为奇数时,11nn
aa+=−,当n为偶数时,122nnaa+=+,且21nnba−=,所以211123211,2,224aababaa=−=====+=,所以121221212222222nnnnnnbaaaab++−−==−===++,∴12nnbb+=,所以nb为以2
为首项,2为公比的等比数列,所以2,nnbn+=N;(2)因为()11321122122212nnnnaaabbb+−−+++=+++==−−,所以()()()()124213211211122nnnnaaaaaabbbnn+−+++=−+−++−=+++−=−−,所以数列na的前2n
项和2224nnSn+=−−;综上,所以2,nnbn+=N,数列na的前2n项和2224nnSn+=−−.例56.(2022·辽宁·抚顺市第二中学三模)已知数列na中,满足()1212,,n
nnaaabakaa++===+对任意*nN都成立,数列na的前n项和为nS.(1)若na是等差数列,求k的值;(2)若1ab==,且1nnaa++是等比数列,求k的值,并求nS.【解析】(
1)若na是等差数列,则对任意*nN,121nnnnaaaa+++−=−,即122nnnaaa++=+,所以()1212nnnaaa++=+,故12k=.(2)因为121aa==且()12nnnakaa++=+得3421111,1aakk
k=−=−−,又1nnaa++是等比数列,则()()()2231234aaaaaa+=++即221122kk=−,得12k=.当12k=时,121,1aa==,1na=,故1nnaa++是以2为首项,公比为1的等比数列,此时na的前n项和
nSn=;当12k=−时,()1212nnnaaa++=−+,即122nnnaaa++=−−,所以()211nnnnaaaa++++=−+,且1220aa+=所以1nnaa++以122aa+=为首项,公比为-1的等比数列,又()32211nnnnnnaaaaaa
++++++=−+=+,所以,当n是偶数时,()()()1234112341nnnnnSaaaaaaaaaaaa−−=++++++=++++++()122naan=+=,当n是奇数时,()23212aaaa+=−+=−
,()()()12341123451nnnnnSaaaaaaaaaaaaa−−=++++++=+++++++()11222nn−=+−=−2,21,2nnnkSnnk−=−==,()*kN综上,当12k=时,nSn=,当12k=−时2,21,2nnnkS
nnk−=−==,()*kN.例57.(2022·湖南·高三阶段练习)已知数列na中,11a=,12nnnaa+=,令2nnba=.(1)求数列nb的通项公式;(2)若222,22,loglognnnnbncnbb+=+为奇
数为偶数,求数列nc的前14项和.【解析】(1)当1n=时,122aa=,又11a=,得22a=,由12nnnaa+=①得1122nnnaa+++=②,①②两式相除可得22nnaa+=,则12222nnnnbaba++==,且122ba==,所以数列nb是以2为首项,2为公比的等比数列
,故2nnb=.(2)当n为奇数时,1222nnnbc−==;当n为偶数时,2lognbn=,222222loglog2nnncnnbbnn+===+−+++.所以数列nc的前14项和为()()1214
13132414ccccccccc+++=+++++++()()()()26122242641614=+++++−+−++−72142=−+−1312=−.例58.(2022·全国·模拟预测)已知数列na满足11a=,1,,2,.nnna
naan+=为奇数为偶数(1)令2nnba=,求1b,2b及nb的通项公式;(2)求数列na的前2n项和2nS.【解析】(1)由题意得211aa==,3222aa==,432aa==,5424aa==,654aa==,121ba==,242ba=
=,364ba==,当2n时,22122122nnnnnbaaab−−−====,又11b=,所以nb是以1为首项,2为公比的等比数列,所以12nnb−=.(2)由(1)知122nna−=,所以12122nnnaa−−==,所以()()212342
121321242nnnnnSaaaaaaaaaaaa−−=++++++=+++++++()()111121212421242221212nnnnn−−+−−=+++++++++=+=−−−.例59.(2022
·全国·高三专题练习)已知数列na的前n项和为nS,且2,,为奇数为偶数=nnnSnn(1)求na的通项公式;(2)设1nnnbaa+=+,求数列nb的前20项和20T.【解析】解:(1)当1n
=时,112aS==当n为奇数,且3n时,12(1)1nnnaSSnnn−=−=−−=+,显然1n=满足;当n为偶数时,12(1)2nnnaSSnnn−=−=−−=−所以1,,2,.nnnann+=−为奇数为偶数(2)()()()2012201223
2021Tbbbaaaaaa=+++=++++++()1220211212aaaaaa=++++−−211212Saa=−−222122260=−−=.例60.(2022·重庆·高三阶段练习)已知数列na的前n项和()2nSnnR=+,
且36a=,正项等比数列nb满足:11ba=,2324bbaa+=+.(1)求数列na和nb的通项公式;(2)若2021nncb=−,求数列nc的前n项和nT.【解析】(1)当2n时,()221(1)121nnnaSSn
nnnn−=−=+−−+−=−+,由36a=,得1=,即2nSnn=+,当1n=时,112aS==,当2n时,2nan=,所以2nan=;设正项等比数列nb的公比为(0)qq,则()21123242,212babbaaqq==+=+=+=,所以260qq+
−=,解得2q=或3q=−(舍),所以2nnb=.(2)()()20212,10202122021,11nnnnncbn−=−=−,所以当10n时,()()1212122021222202120212212nnnnTnnn+−=−+++=−=−+−
,当11n时,()1112112102021222202122404202422021404222nnnnTnTnn+++=−−++=−+−+−+=−+−,即()()1112220212,10.220214042
22,11nnnnnTnn++−++=−+−【方法技巧与总结】(1)分奇偶各自新数列求和(2)要注意处理好奇偶数列对应的项:①可构建新数列;②可“跳项”求和【过关测试】一、单选题1.(2022·全国·高三专题练习)数列
()()121nn−−的前2022项和等于()A.1010−B.2022C.2018−D.2019【答案】B【解析】解:设数列()()121nn−−的前2022项和为2022S,当n为奇数时()()()12121nnn−−=−−,当n为偶
数时()()12121nnn−−=−,所以()2022135791140414043S=−+−+−+++−+()()()()135791140414043=−++−++−+++−+210112022==.故选:B2.(2022·江西·临川一中模拟预测(文))已知
数列na的通项公式为cos1),(nnannS=−为数列的前n项和,2021S=()A.1008B.1009C.1010D.1011【答案】D【解析】解:因为当n为奇数时cos11()n−=,n为偶数时
cos1)1(n−=−,所以()1()cos11nn−−=−,所以()1cos1)1(nnannn−=−=−,所以20211234202020211101020211011S=−+−+−+=−+=;故选
:D3.(2022·四川·射洪中学模拟预测(文))已知首项为1的等差数列na的前n项和为nS,满足202220211202220212SS−=,则202011iiS==()A.20202021B.40402021C.20212022D.40422022【答案】B【解析】由2022202112
02220212SS−=可得:nSn为等差数列,公差为12,首项为11111Sa==,所以()11111222nSnnn=+−=+,则21122nSnn=+,()1211211nSnnnn==−++,所以202
011111111404021212232020202120212021iiS==−+−++−=−=故选:B4.(2022·全国·高三专题练习)己知数列na满足nan=,在1,nn
aa+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【解析】解:数列{}na满足nan=,在n
a,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n
an=,所以()100121311313()(10013)1871782Saaa+=++++−=+=.故选:A.5.(2022·河南·南阳中学高三阶段练习(文))已知数列na满足()2*12nnnaaan++=N,12a=,416a=,数列nb满足2(1)lognnnn
baa=+−,则数列nb的前2021项的和2021S为()A.202222025−B.202221007+C.202221008+D.202221013−【答案】D【解析】因为()2*12nnnaaan++=N,故数列na为等比数列,又142,16aa==,所以2nna=;则22
(1)log22(1)nnnnnnbn=+−=+−;所以122021(222)(1234201920202021)nS=++++−+−++−+−202120222(12)101020212101312−=+−=
−−.故选:D.6.(2022·全国·高三专题练习)已知公比为2的等比数列na满足38a=,记mb为na在区间(0,m(m为正整数)中的项的个数,则数列{}mb的前100项的和100S为()A.360B.480C.600D.100【答案】B【解析】解:因为38a=,2q=
,所以332nnnaaq−==,由于123456722,24,28,216,232,264,2128=======,所以1b对应的区间为(0,1],则10b=;23,bb对应的区间分别为(0,2],(0,3],则231bb==,即有2
个1;4567,,,bbbb对应的区间分别为(0,4],(0,5],(0,6],(0,7],则45672bbbb====,即有22个2;8915,,,bbb对应的区间分别为(0,8],(0,9],,(0,15],则89153bbb====,即有32个3;161731,,,b
bb对应的区间分别为(0,16],(0,17],,(0,31],则1617314bbb====,即有42个4;323363,,,bbb对应的区间分别为(0,32],(0,33],,(0,63],则3233635bbb====L,即有52个5;
6465100,,,bbbL对应的区间分别为(0,64],(0,65],,(0,100],则64651006bbb====L,即有37个6.所以23451001222324252637480S=+++++=.故
选:B7.(2022·全国·高三专题练习)已知数列na满足113a=,21nnnaaa+=+,用x表示不超过x的最大整数,则12201111111aaa+++=+++()A.1B.2C.3D.4【答案】B【解析】因为(
)211nnnnnaaaaa+=+=+,所以()1111111nnnnnaaaaa+==−++,即11111nnnaaa+=−+,所以12201122320120212022021111111111113111aaaaaaaaaaaa+++=−+−++−=−=−+++,由113a=,21nnnaa
a+=+可得249a=,35281a=,469166561a=,210nnnaaa+−=,则数列na是递增数列,20241aa,则202101a,则12201202111132111aaaa+++=−=+++.故选:B.8.(20
22·全国·哈师大附中模拟预测(文))已知数列na满足11242nna−=++++,则数列12nnnaa+的前5项和为()A.131B.163C.3031D.6263【答案】D【解析】因为111124221,21nnnnnaa−++=++++=−
=−,所以()()()()()()1111121212211212121212121nnnnnnnnnnnnaa+++++−−−===−−−−−−−.所以12nnnaa+前5项和为1223561611111111162121212121212121216363−+−
++−=−=−=−−−−−−−−故选:D二、多选题9.(2022·全国·高三专题练习)已知下图的一个数阵,该阵第n行所有数的和记作na,11a=,21112a=++,311111242a=++++,L,数列na的前n项和记作nS,则下列说法正确的是()
A.1342nna−=−B.132nnnaa+−=C.522716S=D.3462nnSn=−+【答案】ABC【解析】解:由题意得:A选项:2121111112222nnna−−=+++++++1111111122341121122nnn−−−−
=+=−−−,故A正确;B选项:1133344222nnnnnaa+−−=−−−=,故B正确;D选项:11112343461212nnnSnn−−=−=−+−
,故D错误;C选项:54322746216S=−+=,故C正确.故选:ABC10.(2022·全国·高三专题练习)已知正项数列na的首项为2,前n项和为nS,且()()11112nnnnnnnaaaaS
aS+++−+++=+,112nnnbaa+=+−,数列nb的前n项和为nT,若16nT,则n的值可以为()A.543B.542C.546D.544【答案】AB【解析】因为()()11112nnnnnnnaaaaSaS+++−+++=+,所以()221121nnnnaaaa++−=−
+,即()()221112nnaa+−−−=,故数列()21na−是首项为()2111a−=,公差为2的等差数列,则()2121nan−=−,则211nan=−+,所以1112121222121nnnnnbaann++−−===+−++−,则()()113153212
121122nTnnn=−+−+++−−=+−,令()1211162n+−,解得2133n+,即544n,故选:AB11.(2022·全国·高三专题练习)我们把221nnF=+(0,1,2,n=L)叫作“费马数”(费马是十七世纪法国数学家).设(
)2log1nnaF=−,1,2,n=,nS表示数列na的前n项和,则使不等式12320212nSSSSn++++−成立的正整数n的值可以是()A.7B.8C.9D.10【答案】CD【解析】221nnF=+(0,1,2,n=L),()2log12nn
naF=−=,*nN,()12122212nnnS+−==−−,()2123412222412nnnSSSSnn+−++++=−=−−−,222242021222025nnnn++−−−.当8n=时,左边1024=,不满足题意;当
9n=时,左边2048=,满足题意,故最小正整数n的值为9.故选:CD.12.(2022·河北·模拟预测)将数列{32}n−与2n的公共项从小到大排列得到数列na,则下列说法正确的有()A.数列na为等差数列B.数列na为等比数列C.14nna+=D.数列
(32)nna−的前n项和为1(1)44nn+−+【答案】BD【解析】数列{32}n−中的项为1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49,52,55,58,61,64,67,…,数列2n中的
项为2,4,8,16,32,64,128,…,∴数列na是首项为4,公比为4的等比数列,∴4nna=;∴(32)(3)4=2−−nnnan,记数列(32)nna−的前n项和为nT,则211444(35)4(32)4−=+++−+−nnnTnn,2
3141444(35)4(32)4+=+++−+−nnnTnn,两式相减:231343(444)(32)4+−=++++−−nnnTn2114(14)43(32)414−+−=+−−−nnn114416(32)4++=+−−−nnn1(33)412+=−−−nn,∴
14(1)4+=+−nnTn.故选:BD三、填空题13.(2022·四川成都·模拟预测(理))杨辉三角,是二项式系数在三角形中的一种几何排列.在欧洲,这个表叫做帕斯卡三角形.帕斯卡(1623-1662)是在1654年发现这一规律的,比杨辉要迟393年,比贾宪迟600年.这是我国数学史上的又一个伟
大成就.其实,中国古代数学家在数学的许多重要领域中处于遥遥领先的地位.中国古代数学史曾经有自己光辉灿烂的篇章,而杨辉三角的发现就是十分精彩的一页.下图的表在我国南宋数学家杨辉1261年所著的《详解九章算法
》一书里就出现了.该表中,从上到下,第n次出现某行所有数都是奇数的行号记为na,比如121,3aa==,则数列na的前10项和为___________.第1行11第2行121第3行1331第4行14641第5行15101051第6行1615201561【答案】2036【解
析】容易发现37a=,415a=,531a=归纳可得21nna=−,故na的前10项和为23101122221022102036++++−=−−=.故答案为:2036.14.(2022·四川省内江市第六中学模拟
预测(理))已知数列na满足12a=,24a=,2(1)3+−=−+nnnaa,则数列na的前20项和为___________.【答案】330【解析】由题意,当n为奇数时,2(1)32nnaa+−=−+=
,所以数列21na−是公差为2,首项为2的等差数列,所以2122(1)2nnan−=+−=,当n为偶数时,2134nnaa+−=+=,所以数列2na是公差为4,首项为4的等差数列,所以244(1)4nnna=+−=,()()20122
013192420......Saaaaaaaaa=+++=+++++++10(220)10(440)(2420)(4840)33022++=+++++++=+=,故答案为:33015.(2022·上海·
模拟预测)设12nCCC、、、、是坐标平面上的一列圆,它们的圆心都在x轴的正半轴上,且都与直线33yx=相切,对每一个正整数n,圆nC都与圆1nC+相互外切,以nr表示圆nC的半径,已知nr为递增数列,若11r=,则数列nnr的前n项和为_
________.【答案】()21314nn−+【解析】33yx=的倾斜角π6=,设圆nC、1nC+与直线33yx=的切点分别为EF、,连接1,nnCECF+,过nC作1nnCDCF+⊥,垂足为D,则1111,nnnnnnnCCrrCDrr++++=+=−∵111112nnnnnnnC
DrrCCrr++++−==+,整理得13nnrr+=数列nr是以首项11r=,公比3q=的等比数列,即11133nnnr−−==∴13nnnnr−=,设数列nnr的前n项和为nS,则有:0121132333.
..3nnnS−++++=123132333...33nnSn++++=两式相减得:()01212131132333...333132nnnnnnnnnS−−+−−++++−=−==−−即(
)21314nnnS−+=故答案为:()21314nn−+.16.(2022·全国·高三专题练习)设数列na的前n项和为nS,已知1222,(1)2nnnaaa−+=+−=,则60S=_______
__.【答案】960【解析】由12(1)2nnnaa−++−=,当n为奇数时,有22nnaa++=;当n为偶数时,22nnaa+−=,∴数列na的偶数项构成以2为首项,以2为公差的等差数列,则()()6
01357575924685860Saaaaaaaaaaaa=+++++++++++++302915230229602=++=,故答案为:960.四、解答题17.(2022·湖北·模拟预测)已知数列na,nb满足10a=
,13b=,且112136nnnaab−−=+,111536nnnbab−−=+.(1)若nnab+为等比数列,求值;(2)在(1)的条件下,求数列na的前n项和nS.【解析】(1)由题1121536nnnnabab−−+++=+∵nnab+
为等比数列,设公比为q则()11nnnnababq−−+=+∴21356qq+=+=,∴21536++=,即220+−=,解得2=−或1=当2=−时,12q=,即()11
1222nnnnabab−−−+=−+又1123ab−+=,∴2nnab−+成以3为首项,以12为公比的等比数列当1=时,1q=即11nnnnabab−−+=+又113ab+=,∴nnab+成以3为首项,以1为公比的等比数列综上:2=−或1=(2)由(1)得11232nnnab−
−+=,3nnab+=∴1112nna−=−∴0121111111112222nnS−=−+−+−++−1111221212nnnn−−=−=+−−
18.(2022·广东·深圳市光明区高级中学模拟预测)已知各项都为正数的数列{}na满足1+32nnnaa+=,11a=.(1)若2nnnba=−,求证:{}nb是等比数列;(2)求数列{}na的前n项和nS.【解析】(1)因为1+32nnnaa
+=所以()111123222nnnnnnnnnbaaab++++=−=−+−=−−=−,因为1=1a,所以1121ba=−=−所以10nnbb+=−所以11nnbb+=−所以nb是首项和公比均为1−的等比数列.
(2)由(1)易得:(1)nnb=−因为()2-1=nnnnba=−所以2(1)nnna=+−所以()()()()1232=1111222nnnS−+−+−++−++++()()()111212212nn−−−−=+−1(1)5222
+−=+−nn19.(2022·山东·肥城市教学研究中心模拟预测)已知数列na为公差不为零的等差数列,其前n项和为nS,1712aa+=,525S=.(1)求数列na的通项公式;(2)令2lognnca=,其中x表示不超过x的最大
整数,求1220ccc+++的值.【解析】(1)设数列na为公差为d,1712aa+=,525S=,∴()111612545252aadad++=+=∴13,1ad==∴数列na的通项公式为2nan=+(2)2nan=+
,则13a=,20162232a=,当()22loglog21nncan==+=,则224n+,可得1n=,当()22loglog22nncan==+=,则428n+,可得26n,当()22log
log23nncan==+=,则8216n+,可得614n,当()22loglog24nncan==+=,则16232n+,可得1431n,此时1420n.所以,1,12,263,6144,1420nn
ncnn==,故1220124384761ccc+++=+++=20.(2022·江西萍乡·三模(理))已知正项数列na的前n项和nS满足:12(N)nnSaan+=−,且123+1
,aaa,成等差数列.(1)求数列na的通项公式;(2)令()()()2221Nloglognnnbnaa++=,求证:数列nb的前n项和34nT.【解析】(1)由题意:()12,nnSaanN+=−,()-1-112,2,NnnSaann+=−
两式相减得到-1=2(2,)nnaannN+,又0na,na是首项为1a,公比为2的等比数列,再由123+1,aaa,成等差数列得,得()2132+1aaa=+,即()11122+14aaa=+,则1=2a,na的通项公式为()2Nnnan+=.(2)由题意知,22211
111()(2)22log2log2nnnbnnnn+===−++1111111111(1)232435112nTnnnn=−+−+−++−+−−++11113111122124212nnnn=+−−=−+++++
3N,4nnT+21.(2022·宁夏·银川一中模拟预测(文))已知数列na是等差数列,nb是等比数列,且22b=,34b=,11ab=,851ab+=.(1)求数列na、nb的通项公式;(2
)设11nnnacb++=,数列nc的前n项和为nS,求nS.【解析】(1)依题意,等比数列nb的公比322bqb==,则有2122nnnbbq−−==,因此,111ab==,由851ab+=得8511
5ab=−=,等差数列na的公差81281aad−==−,1(1)21naandn=+−=−,所以数列na、nb的通项公式分别为:21nan=−,12nnb−=.(2)由(1)知,111222nnnnnanncb−++==
=,则23123412222nnnS−=+++++,于是得23111231222222nnnnnS−−=+++++,两式相减得:23111()11112212122222211222nnnnnnnnSn−−+=+++++−=−=−−,所以1242nnnS−+=−.22.(2
022·浙江·杭师大附中模拟预测)数列na的前n项和为nS,数列nb满足()Nnnbnan=,且数列nb的前n项和为(1)2nnSn−+.(1)求12,aa,并求数列na的通项公式;(2)抽去数列na中点第1项,第4项,第7项,…,第32n
−项,余下的项顺序不变,组成一个新数列nc,数列nc的前n项和为nT,求证:1121153nnTT+.【解析】(1)由题意得12323(1)2nnaaananSn++++=−+,①当1n=时,12a=;当2n=时,1221
222444aaSaaa+=+=++=;当2n时,1231123(1)(2)2(1)nnaaananSn−−++++−=−+−,②①−②得,1(1)(2)2(2)222(2)nnnnnnnnanSnSSnaSan−=−−−
+=+−+=−,当1n=时,12a=,也适合上式,所以()22NnnSan=−,所以1122nnSa−−=−,两式相减得12(2)nnaan−=,所以数列na是以2为首项,2为公比的等比数列,所以2nna=.(2)数列nc为:2
356892,2,2,2,2,2,,所以奇数项是以4为首项,8为公比的等比数列,偶数项是以8为首项,8为公比的等比数列.所以当()21Nnkk=−时,()()122113212422nkkkTccccccccc−−−=+++=+++++++()()()()125313633418
8185812222222181877kkkkk−−−−−=+++++++=+=−−−所以31158121281227777kkknnnTTc++=+=−+=−,所以1128128412812127
75581258125581277kknkkknTT+−−===+−−−,显然1+nnTT是关于k的减函数,所以11235nnTT+;所以当()2Nnkk=时,()()12213212
42nkkkTccccccccc−=+++=+++++++()()()()253136341881812812222222181877kkkkk−−−=+++++++=+=−−−所以3211128124081227777kkknnnTTc+++=+=
−+=−,所以14081240812107771281212812338377kknkkknTT+−−===+−−−,显然1+nnTT是关于k的减函数,所以1101133nnTT+;综上所述,1121153nnTT+.