专题14数列求和综合必刷100题任务一:善良模式(基础)1-30题一、单选题1.已知数列na满足13a,111nnaann,则na()A.14nB.14nC.12nD.12n【答案】B【分析】由1111nnaann,利用累加法得出na.【详解】由题意可得111111nnaannnn,所以21112aa,321123aa,…,1111nnaann,上式累加可得121321nnnaaaaaaaa111111112231nnn,又13a,所以14nan.故选:B.2.已知数列{}na的前n项和为nS,且11a,121()nnaannN,则数列1{}nS的前2020项的和为()A.20202021B.40402021C.40392020D.40412022【答案】B【分析】首先根据已知条件求得na,然后求得nS,利用裂项求和法求得正确答案.【详解】数列{}na的前n项和为nS,且11a,121nnaan,则2132aa.所以2123nnaan,两式相减得:22nnaa,且11a,22a,当n为奇数时,11121122nnaann,当n为偶数时,212222nnaann,所以nan,所以数列na是首项为1,公差为1的等差数列.所以(1)2nnnS,故12112()(1)1nSnnnn,所以121111111112(1)2(1)22311nnTSSSnnn,则2020140402(1)20212021T.故选:B3.数列1,1+2,1+2+22,…,1+2+22+…+2n-1,…的前99项和为()A.2100-101B.299-101C.2100-99D.299-99【答案】A【分析】由数列可知an=1+2+22+…+2n-1=1212n=2n-1,结合分组求和法即可求解.【详解】由数列可知an=1+2+22+…+2n-1=1212n=2n-1,所以,前99项的和为S99=(2-1)+(22-1)+…+(299-1)=2+22+…+299-99=992(12)12-99=2100-101.故选:A4.已知数列na的前n项和nS满足2nSn,记数列11nnaa的前n项和为nT,*nN.则使得20T的值为()A.1939B.2041C.3839D.4041【答案】B【分析】由2nSn,求得21nan,得到11111()22121nnaann,结合裂项法求和,即可求解.【详解】数列na的前n项和nS满足2nSn,当1n时,111aS;当2n时,221(1)21nnnaSSnnn,当1n时,11a适合上式,所以21nan,则111111()(21)(21)22121nnaannnn,所以20111111111120[(1)()()()](1)233557394124141T.故选:B.5.已知数列{an}满足:an+1=an-an-1(n≥2,n∈N*),a1=1,a2=2,Sn为数列{an}的前n项和,则S2021=()A.3B.2C.1D.0【答案】C【分析】根据递推关系式得出数列是周期为6的周期数列,利用周期性即可求解.【详解】∵an+1=an-an-1,a1=1,a2=2,∴a3=1,a4=-1,a5=-2,a6=-1,a7=1,a8=2,…,故数列{an}是周期为6的周期数列,且每连续6项的和为0,故S2021=336×0+a2017+a2018+…+a2021=a1+a2+a3+a4+a5=1+2+1+(-1)+(-2)=1.故选:C.6.正项数列{}na满足11a,211(2)30(1,)nnnnaaaannN,则133520192021111aaaaaaL()A.12003534B.10106061C.12202021D.20205461【答案】B【分析】对211(2)30(1,)nnnnaaaannN化简可得13nnaa,从而可得数列{}na是等差数列,首项为1,公差为3,求出通项na,则可得212111111()(65)(61)66561nnaannnn,然后利用裂项求和法计算【详解】211(2)30(1,)nnnnaaaannN,1(1)[(3)]0nnnaaa,0na,13nnaa,数列{}na是等差数列,首项为1,公差为3,13(1)32nann.212111111()(65)(61)66561nnaannnn,133520192021111111111111010[(1)()()](1)6771360556061660616061aaaaaa.故选:B.7.化简221(1)2(2)2222nnnSnnn的结果是()A.122nnB.122nnC.22nnD.122nn【答案】D【分析】用错位相减法求和.【详解】221(1)2(2)2222nnnSnnn,(1)23122(1)2(2)2222nnnSnnn,(2)(2)-(1)得:21112(12)2222222212nnnnnnSnnnn.故选:D.8.已知数列na中,*111,34(,2)nnaaanNn,求数列na的前n项和nS为()A.13232nnnSB.13232nnnSC.13432nnnSD.1332nnS【答案】C【分析】根据题意化简得到123(2)nnaa,得到数列2na构成首项为3,公比为3的等比数列,求得32nna,结合等比数列和等差数列的求和公式,即可求解.【详解】由题意,数列na中,*111,34(,2)nnaaanNn,可得112363(2)nnnaaa,即1232nnaa,且123a,所以数列2na构成首项为3,公比为3的等比数列,所以23nna,即32nna,则数列na的前n项和2333222nnS1131333343221322nnnnnn.故选:C.9.等比数列na中,12a,2q=,数列111nnnnabaa,nb的前n项和为nT,则10T的值为()A.40944095B.20462047C.10221023D.510511【答案】B【分析】先求出na,从而可得1112112212121nnnnnnb,然后利用裂项相消求和法可求出10T【详解】由题意得2nna,所以1121121212121nnnnnnb,所以10223101111111111120461212121212121212047T.故选:B10.已知数列na的前n项和nS满足2nSn,记数列11nnaa的前n项和为nT,*nN.则使得2041nT成立的n的最大值为()A.17B.18C.19D.20【答案】C【分析】根据1nnnaSS求na通项公式,注意讨论1n、2n并判断是否可合并,再应用裂项法求nT,最后根据不等式求n的最大值即可.【详解】当1n时,111aS;当2n时,221(1)21nnnaSSnnn;而12111a也符合21nan,∴21nan,*nN.又11111()22121nnaann,∴11111111(1...)(1)2335212122121nnTnnnn,要使2041nT,即202141nn,得20n且*nN,则n的最大值为19.故选:C.第II卷(非选择题)二、填空题11.数列na是首项和公差都为1的等差数列,其前n项和为nS,若nT是数列12nS的前n项和,则99T______【答案】99100/0.99.【分析】首先写出等差数列前n项和nS,则有1121nSnn,再应用裂项相消法求99T.【详解】由题意:nan,故12nnnS,于是1111211nSnnnn,∴9911111199...122399100100T.故答案为:99100.12.已知数列na的通项公式*21log()2nnanNn,设其前n项和为nS,则使3nS成立的最小的自然n为__________.【答案】14【分析】先利用其通项公式以及对数函数的运算公式求出22log2nSn.再利用对数的运算性质解不等式3nS„即可求出对应的自然数.【详解】解:因为21log(*)2nnanNn,所以123nnSaaaa22222341loglogloglog3452nn22341log3452nn22log2n.32223log321422nSnnn剟剠.故答案为:14.13.已知数列na满足*2nnaannN,则na的前20项和20S________.【答案】95【分析】利用分组求和法以及等差数列的前n项和公式即可求出结果.【详解】因为*2nnaannN,则*131nnaannN,所以*12321nnnnaaaannN所以201220Saaa123417181920aaaaaaaa21125129121312171215913175117525295,故答案为:95.14.已知正项数列na满足11a,2+11(2)30,(2,)nnnnaaaannN,则122320002021111aaaaaa___________.【答案】20206061【分析】化简数列的递推关系式,得到13nnaa,结合等差数列的通项公式,求得32nan,可得11111()33231nnaann,利用裂项法,即可求解.【详解】由题意,正项数列na满足11a,2+11(2)30,(2,)nnnnaaaannN,可得1(1)(3)0nnnaaa,因为0na,可得13nnaa,所以数列na是首项为1,公差为3的等差数列,所以13(1)32nann,则111111()(32)(31)33231nnaannnn所以122320202021111111111(1)344760586061aaaaaa112020(1).360616061故答案为:20206061.15.设数列na满足12(1)nnaan,*nN,12a,则数列(1)nna的前50项和是________.【答案】1300【分析】利用累加法可求得数列na的通项公式(1)nann,再并项求和求解前50项和即可.【详解】因为12(1)nnaan,*nN,且12a,故2n时,214aa,326aa