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1、專題能力訓(xùn)練12數(shù)列的通項(xiàng)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(xiàng)和,則S2 016=()A.2 016B.-2 016C.3 024D.-3 0242.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=n2+n,數(shù)列bn滿足bn=1anan+1(nN*),Tn是數(shù)列bn的前n項(xiàng)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(xiàng)和為Sn,a1=1,a2=2,且對(duì)于任意n1,nN*,滿足Sn+1+Sn-1=2(Sn+1),則S10的值為()A.90B.91C.

2、96D.1004.設(shè)數(shù)列an的前n項(xiàng)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(gè)新數(shù)列a1,a2-a1,a3-a2,an-an-1,此數(shù)列是首項(xiàng)為1,公比為13的等比數(shù)列,則數(shù)列an的通項(xiàng)公式為()A.an=323213n,nN*B.an=32+3213n,nN*C.an=1,n=1,32+3213n,n2,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

3、(nN*),則數(shù)列an的通項(xiàng)公式an=.8.(2019福建廈門高二檢測(cè),15)已知數(shù)列an滿足3a1+32a2+33a3+3nan=2n+1,則an的通項(xiàng)公式為.9.設(shè)數(shù)列an的前n項(xiàng)和為Sn.已知S2=4,an+1=2Sn+1,nN*.(1)求通項(xiàng)公式an;(2)求數(shù)列|an-n-2|的前n項(xiàng)和.10.(2019廣東汕頭一模,17)已知數(shù)列an的前n項(xiàng)和為Sn,且2Sn=nan+2an-1.(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列1an2的前n項(xiàng)和為Tn,證明:Tnn+2,故bn=3n-1-n-2,n3.設(shè)數(shù)列bn的前n項(xiàng)和為Tn,則T1=2,T2=3.當(dāng)n3時(shí),Tn=3+9(1-3n-2)

4、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(dāng)n=1時(shí),2S1=a1+2a1-1,則a1=1.當(dāng)n2時(shí),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)24n(n+1)=41n-1n+1.所以Tn=422+432

5、+442+4(n+1)2412+423+434+4n(n+1)=41-12+12-13+13-14+1n-1n+1=41-1n+14.11.解(1)由a1=2,an+1=2an,得an=2n(nN*).由題意知,當(dāng)n=1時(shí),b1=b2-1,故b2=2.當(dāng)n2時(shí),1nbn=bn+1-bn,整理得bn+1n+1=bnn,所以bn=n(nN*).(2)由(1)知anbn=n2n,因此Tn=2+222+323+n2n,2Tn=22+223+324+n2n+1,所以Tn-2Tn=2+22+23+2n-n2n+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

6、總價(jià)為n910n-1萬元.這堆貨物的總價(jià)W=1+2910+39102+n910n-1,910W=1910+29102+39103+n910n,兩式相減,得110W=-n910n+1+910+9102+9103+910n-1=-n910n+1-910n1-910=-n910n+10-10910n,則W=-10n910n+100-100910n=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(xiàng)為1S1=-1,公差為d=-1,1Sn=-n,Sn=-1n.14.(1)證明由條件,對(duì)

7、任意nN*,有an+2=3Sn-Sn+1+3,因而對(duì)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(duì)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(xiàng)a1=1,公比為3的等比數(shù)列;數(shù)列a2n是首項(xiàng)a2=2,公比為3的等比數(shù)列.因此a2n-1=3n-1,a2n=23n-1.于是S2n=a1+a2+a2n=(a1+a3+a2n-1)+(a2+a4+a2n)=(1+3+3n-1)

8、+2(1+3+3n-1)=3(1+3+3n-1)=3(3n-1)2,從而S2n-1=S2n-a2n=3(3n-1)2-23n-1=32(53n-2-1).綜上所述,Sn=32(53n-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=a11-q61-q=63,知q-1,所以a11-261-2=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(xiàng)為12,公差為1的等差數(shù)列.設(shè)數(shù)列(

9、-1)nbn2的前n項(xiàng)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=22n-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=-n22n-1,n為奇數(shù),n22n-2,n為偶數(shù),c2n-1+c2n=-(2n-1)222n-2+(2n)222n-2=(4n-1)4n-1,T2n=340+741+(4n-1)4n-1,4T2n=341+742+(4n-5)4n-1+(4n-1)4n,-3T2n=3+441+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.