專題能力訓(xùn)練12數(shù)列的通項與求和一、能力突破訓(xùn)練1.已知數(shù)列an是等差數(shù)列,a1=tan 225°,a5=13a1,設(shè)Sn為數(shù)列(-1)nan的前n項和,則S2 016=()A.2 016B.-2 016C.3 024D.-3 0242.已知數(shù)列an的前n項和為Sn,且Sn=n2+n,數(shù)列bn滿足bn=1ana n+1(nN*),Tn是數(shù)列bn的前n項和,則T9等于()A.919B.1819C.2021D.9403.(2019河北衡水中學(xué)二調(diào),6)已知數(shù)列an的前n項和為Sn,a1=1,a2=2,且對于任意n>1,nN*,滿足Sn+1+Sn-1=2(Sn+1),則S10的值為()A.90B.91C.96D.1004.設(shè)數(shù)列an的前n項和為Sn,且a1=1,Sn+nan為常數(shù)列,則an=()A.13n-1B.2n(n+1)C.1(n+1)(n+2)D.5-2n35.已知數(shù)列an,構(gòu)造一個新數(shù)列a1,a2-a1,a3-a2,an-an-1,此數(shù)列是首項為1,公比為13的等比數(shù)列,則數(shù)列an的通項公式為()A.an=3232×13n,nN*B.an=32+32×13n,nN*C.an=1,n=1,32+32×13n,n>2,且nN*D.an=1,nN*6.若數(shù)列an滿足an+1=11-an,a11=2,則a1=. 7.(2019云南師范大學(xué)附中高三月考,15)在數(shù)列an中,a2=5,an+1-an=2n(nN*),則數(shù)列an的通項公式an=. 8.(2019福建廈門高二檢測,15)已知數(shù)列an滿足3a1+32a2+33a3+3nan=2n+1,則an的通項公式為. 9.設(shè)數(shù)列an的前n項和為Sn.已知S2=4,an+1=2Sn+1,nN*.(1)求通項公式an;(2)求數(shù)列|an-n-2|的前n項和.10.(2019廣東汕頭一模,17)已知數(shù)列an的前n項和為Sn,且2Sn=nan+2an-1.(1)求數(shù)列an的通項公式;(2)若數(shù)列1an2的前n項和為Tn,證明:Tn<4.11.已知數(shù)列an和bn滿足a1=2,b1=1,an+1=2an(nN*),b1+12b2+13b3+1nbn=bn+1-1(nN*).(1)求an與bn;(2)記數(shù)列anbn的前n項和為Tn,求Tn.二、思維提升訓(xùn)練12.(2019安徽合肥第二次質(zhì)檢,11)“垛積術(shù)”(隙積術(shù))是由北宋科學(xué)家沈括在夢溪筆談中首創(chuàng),南宋數(shù)學(xué)家楊輝、元代數(shù)學(xué)家朱世杰豐富和發(fā)展的一類數(shù)列求和方法,有茭草垛、方垛、芻童垛、三角垛等.如圖,某倉庫中部分貨物堆放成“菱草垛”:自上而下,第一層1件,以后每一層比上一層多1件,最后一層是n件.已知第一層貨物的單價是1萬元,從第二層起,貨物的單價是上一層單價的910.若這堆貨物的總價是100-200910n萬元,則n的值為()A.7B.8C.9D.1013.設(shè)Sn是數(shù)列an的前n項和,且a1=-1,an+1=SnSn+1,則Sn=. 14.設(shè)數(shù)列an的前n項和為Sn,已知a1=1,a2=2,且an+2=3Sn-Sn+1+3,nN*.(1)證明:an+2=3an;(2)求Sn.15.已知an是等比數(shù)列,前n項和為Sn(nN*),且1a11a2=2a3,S6=63.(1)求an的通項公式;(2)若對任意的nN*,bn是log2an和log2an+1的等差中項,求數(shù)列(-1)nbn2的前2n項和.16.(2019湖南湘西四校聯(lián)考,17)已知數(shù)列an,bn,Sn為數(shù)列an的前n項和,a1=2b1,Sn=2an-2,nbn+1-(n+1)bn=n2+n.(1)求數(shù)列an的通項公式;(2)證明:數(shù)列bnn為等差數(shù)列;(3)若cn=-anbn2,n為奇數(shù),anbn4,n為偶數(shù),求數(shù)列cn的前2n項和.專題能力訓(xùn)練12數(shù)列的通項與求和一、能力突破訓(xùn)練1.C解析a1=tan225°=1,a5=13a1=13,則公差d=a5-a15-1=13-14=3,an=3n-2.又(-1)nan=(-1)n(3n-2),S2016=(a2-a1)+(a4-a3)+(a6-a5)+(a2014-a2013)+(a2016-a2015)=1008d=3024.2.D解析數(shù)列an的前n項和為Sn,且Sn=n2+n,當n=1時,a1=2;當n2時,an=Sn-Sn-1=2n,an=2n(nN*),bn=1anan+1=12n(2n+2)=141n-1n+1,T9=141-12+12-13+19-110=14×1-110=940.3.B解析Sn+1+Sn-1=2(Sn+1),Sn+1-Sn=Sn-Sn-1+2,an+1-an=2.當n2時,數(shù)列an是等差數(shù)列,公差為2.又a1=1,a2=2,S10=1+9×2+9×82×2=91.4.B解析數(shù)列an的前n項和為Sn,且a1=1,S1+1×a1=1+1=2.Sn+nan為常數(shù)列,Sn+nan=2.當n2時,(n+1)an=(n-1)an-1,從而a2a1·a3a2·a4a3··anan-1=13·24··n-1n+1,an=2n(n+1).當n=1時上式成立,an=2n(n+1).5.A解析因為數(shù)列a1,a2-a1,a3-a2,an-an-1,是首項為1,公比為13的等比數(shù)列,所以an-an-1=13n-1,n2.所以當n2時,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=3232×13n.又當n=1時,an=3232×13n=1,則an=3232×13n,nN*.6.12解析由a11=2及an+1=11-an,得a10=12.同理a9=-1,a8=2,a7=12,.所以數(shù)列an是周期為3的數(shù)列.所以a1=a10=12.7.2n+1解析由題意可得an-an-1=2n-1,an-1-an-2=2n-2,a2-a1=2,利用累加法,得an-a1=2(2n-1-1)2-1=2n-2.又a2-a1=2,a2=5,則a1=3,所以an=2n+1.8.an=1,n=1,13×23n-1,n2解析當n=1時,由3a1=2+1=3,得a1=1;當n2時,由3a1+32a2+33a3+3nan=2n+1,得3a1+32a2+33a3+3n-1an-1=2n-1+1,兩式相減,得3nan=2n-1,即an=13×23n-1,故an=1,n=1,13×23n-1,n2.9.解(1)由題意,得a1+a2=4,a2=2a1+1,則a1=1,a2=3.又當n2時,由an+1-an=(2Sn+1)-(2Sn-1+1)=2an,得an+1=3an.所以,數(shù)列an的通項公式為an=3n-1,nN*.(2)設(shè)bn=|3n-1-n-2|,nN*,b1=2,b2=1.當n3時,由于3n-1>n+2,故bn=3n-1-n-2,n3.設(shè)數(shù)列bn的前n項和為Tn,則T1=2,T2=3.當n3時,Tn=3+9(1-3n-2)1-3(n+7)(n-2)2=3n-n2-5n+112,所以Tn=2,n=1,3n-n2-5n+112,n2,nN*.10.(1)解當n=1時,2S1=a1+2a1-1,則a1=1.當n2時,2Sn=nan+2an-1,2Sn-1=(n-1)an-1+2an-1-1,-,得2an=nan-(n-1)an-1+2an-2an-1,即nan=(n+1)an-1,所以ann+1=an-1n,且a12=12,所以數(shù)列ann+1為常數(shù)列,ann+1=12,即an=n+12(nN*).(2)證明由(1)得an=n+12,所以1an2=4(n+1)2<4n(n+1)=41n-1n+1.所以Tn=422+432+442+4(n+1)2<41×2+42×3+43×4+4n(n+1)=41-12+12-13+13-14+1n-1n+1=41-1n+1<4.11.解(1)由a1=2,an+1=2an,得an=2n(nN*).由題意知,當n=1時,b1=b2-1,故b2=2.當n2時,1nbn=bn+1-bn,整理得bn+1n+1=bnn,所以bn=n(nN*).(2)由(1)知anbn=n·2n,因此Tn=2+2·22+3·23+n·2n,2Tn=22+2·23+3·24+n·2n+1,所以Tn-2Tn=2+22+23+2n-n·2n+1.故Tn=(n-1)2n+1+2(nN*).二、思維提升訓(xùn)練12.D解析由題意,得第n層貨物的總價為n·910n-1萬元.這堆貨物的總價W=1+2×910+3×9102+n·910n-1,910W=1×910+2×9102+3×9103+n·910n,兩式相減,得110W=-n·910n+1+910+9102+9103+910n-1=-n·910n+1-910n1-910=-n·910n+10-10·910n,則W=-10n·910n+100-100·910n=100-200910n,解得n=10.13.-1n解析由an+1=Sn+1-Sn=SnSn+1,得1Sn1Sn+1=1,即1Sn+11Sn=-1,則1Sn為等差數(shù)列,首項為1S1=-1,公差為d=-1,1Sn=-n,Sn=-1n.14.(1)證明由條件,對任意nN*,有an+2=3Sn-Sn+1+3,因而對任意nN*,n2,有an+1=3Sn-1-Sn+3.兩式相減,得an+2-an+1=3an-an+1,即an+2=3an,n2.又a1=1,a2=2,所以a3=3S1-S2+3=3a1-(a1+a2)+3=3a1,故對一切nN*,an+2=3an.(2)解由(1)知,an0,所以an+2an=3,于是數(shù)列a2n-1是首項a1=1,公比為3的等比數(shù)列;數(shù)列a2n是首項a2=2,公比為3的等比數(shù)列.因此a2n-1=3n-1,a2n=2×3n-1.于是S2n=a1+a2+a2n=(a1+a3+a2n-1)+(a2+a4+a2n)=(1+3+3n-1)+2(1+3+3n-1)=3(1+3+3n-1)=3(3n-1)2,從而S2n-1=S2n-a2n=3(3n-1)2-2×3n-1=32(5×3n-2-1).綜上所述,Sn=32(5×3n-32-1),n是奇數(shù),32(3n2-1),n是偶數(shù).15.解(1)設(shè)數(shù)列an的公比為q.由已知,有1a11a1q=2a1q2,解得q=2或q=-1.又由S6=a1·1-q61-q=63,知q-1,所以a1·1-261-2=63,得a1=1.所以an=2n-1.(2)由題意,得bn=12(log2an+log2an+1)=12(log22n-1+log22n)=n-12,即bn是首項為12,公差為1的等差數(shù)列.設(shè)數(shù)列(-1)nbn2的前n項和為Tn,則T2n=(-b12+b22)+(-b32+b42)+(-b2n-12+b2n2)=b1+b2+b3+b4+b2n-1+b2n=2n(b1+b2n)2=2n2.16.(1)解Sn=2an-2,Sn-1=2an-1-2(n2),an=2an-2an-1(n2),an=2an-1(n2),數(shù)列an是以2為公比的等比數(shù)列.又a1=S1=2a1-2,a1=2,an=2·2n-1=2n.(2)證明nbn+1-(n+1)bn=n2+n=n(n+1),bn+1n+1bnn=1,數(shù)列bnn是以1為公差的等差數(shù)列.(3)解b1=1,由(2)知bnn=b1+(n-1)×1=n,bn=n2,cn=-n2·2n-1,n為奇數(shù),n2·2n-2,n為偶數(shù),c2n-1+c2n=-(2n-1)222n-2+(2n)222n-2=(4n-1)·4n-1,T2n=3×40+7×41+(4n-1)·4n-1,4T2n=3×41+7×42+(4n-5)·4n-1+(4n-1)·4n,-3T2n=3+4×41+42+4n-1-(4n-1)·4n=3+16×(1-4n-1)1-4-(4n-1)·4n=4n+1-73-(4n-1)·4n=-4n(12n-3)+4n+1-73,T2n=4n(12n-7)+79.