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考點(diǎn)規(guī)范練30數(shù)列求和基礎(chǔ)鞏固組1.數(shù)列112,314,518,7116,(2n-1)+12n,的前n項(xiàng)和Sn的值等于()A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n答案A解析該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n.2.(2017浙江金華十校3月聯(lián)考)在數(shù)列an中,an+1-an=2,Sn為an的前n項(xiàng)和.若S10=50,則數(shù)列an+an+1的前10項(xiàng)和為()A.100B.110C.120D.130答案C解析數(shù)列an+an+1的前10項(xiàng)和為a1+a2+a2+a3+a10+a11=2(a1+a2+a10)+a11-a1=2S10+102=120.故選C.3.(2017浙江五校聯(lián)考)已知數(shù)列5,6,1,-5,該數(shù)列的特點(diǎn)是從第二項(xiàng)起,每一項(xiàng)都等于它的前后兩項(xiàng)之和,則這個(gè)數(shù)列的前16項(xiàng)之和S16等于()A.5B.6C.7D.16答案C解析根據(jù)題意這個(gè)數(shù)列的前8項(xiàng)分別為5,6,1,-5,-6,-1,5,6,發(fā)現(xiàn)從第7項(xiàng)起,數(shù)字重復(fù)出現(xiàn),所以此數(shù)列為周期數(shù)列,且周期為6,前6項(xiàng)和為5+6+1+(-5)+(-6)+(-1)=0.又因?yàn)?6=26+4,所以這個(gè)數(shù)列的前16項(xiàng)之和S16=20+5+6+1-5=7.故選C.4.(2018浙江稽陽聯(lián)考)已知Sn是等比數(shù)列an的前n項(xiàng)和,若存在mN*,滿足S2mSm=9,a2mam=5m+1m-1,則數(shù)列an的公比為()A.-2B.2C.-3D.3答案B解析設(shè)公比為q,若q=1,則S2mSm=2,與題中條件矛盾,故q1.S2mSm=a1(1-q2m)1-qa1(1-qm)1-q=qm+1=9,qm=8.a2mam=a1q2m-1a1qm-1=qm=8=5m+1m-1.m=3.q3=8.q=2.故選B.5.(2017全國(guó)高考)等差數(shù)列an的首項(xiàng)為1,公差不為0.若a2,a3,a6成等比數(shù)列,則an前6項(xiàng)的和為()A.-24B.-3C.3D.8答案A解析設(shè)等差數(shù)列的公差為d,則d0,a32=a2a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=61+652(-2)=-24,故選A.6.(2017浙江杭州模擬)已知等差數(shù)列an滿足a3=7,a5+a7=26,bn=1an2-1(nN*),數(shù)列bn的前n項(xiàng)和為Sn,則S100的值為.答案n4(n+1)解析設(shè)等差數(shù)列an的公差為d,a3=7,a5+a7=26,a1+2d=7,2a1+10d=26,解得a1=3,d=2.an=3+2(n-1)=2n+1.bn=1an2-1=14n2+4n=141n-1n+1.S100=141-12+12-13+1n-1n+1=141-1n+1=n4(n+1).7.(2018浙江金華十校模擬)已知公差不為零的等差數(shù)列an中,a1=1,且a2,a5,a14成等比數(shù)列,an的前n項(xiàng)和為Sn,bn=(-1)nSn,則數(shù)列bn的前2n項(xiàng)和T2n=.答案n(2n+1)解析a1=1,an是等差數(shù)列,a2,a5,a14成等比數(shù)列,(1+d)(1+13d)=(1+4d)2,解得d=2.an=a1+(n-1)d=2n-1,Sn=na1+n(n-1)2d=n2.bn=(-1)nSn=(-1)nn2,數(shù)列bn的前n項(xiàng)和Tn=(-12+22)+(-32+42)+-(2n-1)2+(2n)2=3+7+4n-1=n(2n+1).故答案為n(2n+1).8.有窮數(shù)列1,1+2,1+2+4,1+2+4+2n-1所有項(xiàng)的和為.答案2n+1-2-n解析由題意知所求數(shù)列的通項(xiàng)為1-2n1-2=2n-1,故由分組求和法及等比數(shù)列的求和公式可得和為2(1-2n)1-2-n=2n+1-2-n.能力提升組9.已知函數(shù)f(x)=xa的圖象過點(diǎn)(4,2),令an=1f(n+1)+f(n),nN*,記數(shù)列an的前n項(xiàng)和為Sn,則S2 017=()A.2016-1B.2017-1C.2018-1D.2018+1答案C解析由f(4)=2得4a=2,解得a=12,則f(x)=x12.故an=1f(n+1)+f(n)=1n+1+n=n+1-n,S2017=a1+a2+a3+a2017=(2-1)+(3-2)+(4-3)+(2018-2017)=2018-1.10.已知等比數(shù)列an的各項(xiàng)都為正數(shù),且當(dāng)n3時(shí),a4a2n-4=102n,則數(shù)列l(wèi)g a1,2lg a2,22lg a3,23lg a4,2n-1lg an,的前n項(xiàng)和Sn等于()A.n2nB.(n-1)2n-1-1C.(n-1)2n+1D.2n+1答案C解析等比數(shù)列an的各項(xiàng)都為正數(shù),且當(dāng)n3時(shí),a4a2n-4=102n,an2=102n,即an=10n.2n-1lgan=2n-1lg10n=n2n-1.Sn=1+22+322+n2n-1,2Sn=12+222+323+n2n,-,得-Sn=1+2+22+2n-1-n2n=2n-1-n2n=(1-n)2n-1.Sn=(n-1)2n+1.11.在數(shù)列an中,an+1+(-1)nan=2n-1,則數(shù)列an的前12項(xiàng)和等于()A.76B.78C.80D.82答案B解析因?yàn)閍n+1+(-1)nan=2n-1,所以a2-a1=1,a3+a2=3,a4-a3=5,a5+a4=7,a6-a5=9,a7+a6=11,a11+a10=19,a12-a11=21.所以a1+a3=2,a4+a2=8,a12+a10=40.所以從第一項(xiàng)開始,依次取兩個(gè)相鄰奇數(shù)項(xiàng)的和都等于2,從第二項(xiàng)開始,依次取兩個(gè)相鄰偶數(shù)項(xiàng)的和構(gòu)成以8為首項(xiàng),以16為公差的等差數(shù)列,將以上各式相加可得S12=a1+a2+a3+a12=(a1+a3)+(a5+a7)+(a9+a11)+(a2+a4)+(a6+a8)+(a10+a12)=32+8+24+40=78.12.(2017浙江杭州學(xué)軍中學(xué)測(cè)試改編)設(shè)Sn是數(shù)列an的前n項(xiàng)和,已知a1=3,an+1=2Sn+3(nN*),bn=(2n-1)an,則數(shù)列bn的前n項(xiàng)和Tn為()A.(n-1)3n+1B.(n-1)3n+1+3C.(n-1)3n+3D.n3n+1+3答案B解析當(dāng)n2時(shí),由an+1=2Sn+3得an=2Sn-1+3,兩式相減,得an+1-an=2Sn-2Sn-1=2an,an+1=3an,an+1an=3.當(dāng)n=1時(shí),a1=3,a2=2S1+3=2a1+3=9,則a2a1=3.數(shù)列an是以a1=3為首項(xiàng),公比為3的等比數(shù)列.an=33n-1=3n.bn=(2n-1)an=(2n-1)3n,Tn=13+332+533+(2n-1)3n,3Tn=132+333+534+(2n-1)3n+1,-,得-2Tn=13+232+233+23n-(2n-1)3n+1=3+2(32+33+3n)-(2n-1)3n+1=3+232(1-3n-1)1-3-(2n-1)3n+1=-6-(2n-2)3n+1.Tn=(n-1)3n+1+3.13.(2017課標(biāo)高考)幾位大學(xué)生響應(yīng)國(guó)家的創(chuàng)業(yè)號(hào)召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動(dòng).這款軟件的激活碼為下面數(shù)學(xué)問題的答案:已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一項(xiàng)是20,接下來的兩項(xiàng)是20,21,再接下來的三項(xiàng)是20,21,22,依此類推,求滿足如下條件的最小整數(shù)N:N100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()A.440B.330C.220D.110答案A解析設(shè)數(shù)列的首項(xiàng)為第1組,接下來兩項(xiàng)為第2組,再接下來三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)2.第n組的和為1-2n1-2=2n-1,前n組總共的和為2(1-2n)1-2-n=2n+1-2-n.由題意,N100,令n(1+n)2100,得n14且nN*,即N出現(xiàn)在第13組之后.若要使最小整數(shù)N滿足:N100且前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(kN*,n14),所以k=log2(n+3),解得n=29,k=5.所以N=29(1+29)2+5=440.故選A.14.(2018浙江長(zhǎng)興模擬)已知數(shù)列an的前n項(xiàng)和Sn滿足Sn+1=kSn+2,a1=2,a2=1,則k=,Sn=.答案1241-12n解析S2=kS1+2,a1+a2=ka1+2.又a1=2,a2=1,2+1=2k+2.k=12.Sn+1=12Sn+2,當(dāng)n2時(shí),Sn=12Sn-1+2,-,得an+1=12an(n2).又a2=12a1,易見an0(nN*),an+1an=12(nN*).數(shù)列an是等比數(shù)列,公比為12,Sn=21-12n1-12=41-12n.15.“斐波那契數(shù)列”是數(shù)學(xué)史上一個(gè)著名數(shù)列,在斐波那契數(shù)列an中,a1=1,a2=1,an+2=an+1+an(nN*),則a7=;若a2 018=m,則數(shù)列an的前2 016項(xiàng)和是(用m表示).答案13m-1解析a1=1,a2=1,an+2=an+1+an(nN*),a3=1+1=2,同理可得:a4=3,a5=5,a6=8,則a7=13.a1=1,a2=1,an+an+1=an+2(nN*),a1+a2=a3,a2+a3=a4,a3+a4=a5,a2015+a2016=a2017,a2016+a2017=a2018.以上累加,得a1+2a2+2a3+2a4+2a2016+a2017=a3+a4+a2018,a1+a2+a3+a4+a2016=a2018-a2=m-1.16.(2018浙江慈溪高三上期中)若數(shù)列an滿足an+1+(-1)nan=2n-1,其前n項(xiàng)和為Sn,則(1)a1+a3+a5+a99=;(2)S4n=.答案(1)50(2)8n2+2n解析(1)an+1+(-1)nan=2n-1,a2n+1+a2n=4n-1,a2n-a2n-1=4n-3.兩式相減得a2n+1+a2n-1=2,則a3+a1=2,a7+a5=2,a99+a97=2,a1+a3+a5+a99=252=50.(2)由(1)得a3=2-a1,a2n+3+a2n+1=2,a2n+3=2-a2n+1=2-(2-a2n-1)=a2n-1(nN*).當(dāng)n=2k(kN*)時(shí),a4k+3=a4k-1=a3=2-a1;當(dāng)n=2k-1(kN*)時(shí),a4k+1=a4k-3=a1.由已知可得a4k-1+a4k-2=8k-5,a4k-a4k-1=8k-3(kN*).a4k-2=8k-5-a4k-1=8k-7+a1,a4k=8k-3+a4k-1=8k-1-a1.an=a1,n=4k-3,2n-3+a1,n=4k-2,2-a1,n=4k-1,2n-1-a1,n=4k(kN*).設(shè)bn=a4n-3+a4n-2+a4n-1+a4n=16n-6(nN*),則數(shù)列bn為首項(xiàng)為10,公差為16的等差數(shù)列.S4n=b1+b2+bn=10n+16n(n-1)2=8n2+2n.17.(2018浙江名校新高考研究聯(lián)盟第一次聯(lián)考)已知數(shù)列an的前n項(xiàng)和為Sn,且滿足Sn=a2n+bn-1(a,bR,nN*).(1)當(dāng)a=1,b=1時(shí),求數(shù)列Sn的前n項(xiàng)和Tn;(2)若an是等比數(shù)列,證明:a2S1S2+a3S2S3+an+1SnSn+11.(1)解當(dāng)a=1,b=1時(shí),Sn=2n+n-1,Tn=S1+S2+Sn=21+1-1+22+2-1+2n+n-1=(21+22+2n)+(1+2+n)-n=2n+1-2+12n(n+1)-n=2n+1+12n(n-1)-2.(2)證明當(dāng)n=1時(shí),a1=S1=2a+b-1,當(dāng)n2時(shí),an=Sn-Sn-1=2n-1a+b,要使得數(shù)列an成等比數(shù)列,則b=0,此時(shí)an=2n-1a,且需滿足當(dāng)n=1時(shí),a1=2a+b-1=a,即a=1,此時(shí):Sn=2n-1,an=2n-1,an+1SnSn+1=2n(2n-1)(2n+1-1)=12n-1-12n+1-1.故a2S1S2+a3S2S3+an+1SnSn+1=2(2-1)(22-1)+22(22-1)(23-1)+2n(2n-1)(2n+1-1)=12-1-122-1+122-1-123-1+12n-1-12n+1-1=1-12n+1-10,即M1M2;當(dāng)n2時(shí),Mn+1-MnM3M4.(Mn)max=M2=1013+14-1=296.故數(shù)列S2n-Sn的最大值為S4-S2=296.