第二講數(shù)列求和及綜合應(yīng)用(40分鐘70分)一、選擇題(每小題5分,共25分)1.在數(shù)列an中,an+1=2an-1,a3=2,設(shè)其前n項(xiàng)和為Sn,則S6=()A.874B.634C.15D.27【解析】選A.因?yàn)閍n+1=2an-1,所以an+1-1=2(an-1),所以an-1是以2為公比的等比數(shù)列,所以an-1=(a1-1)2n-1,因?yàn)閍3=2,所以a1=54,所以an=1+2n-3,所以S6=6+634=874.2.(2018廣東省化州市二模)已知有窮數(shù)列an中,n=1,2,3,729,且an=(2n-1)(-1)n+1,從數(shù)列an中依次取出a2,a5,a14,構(gòu)成新數(shù)列bn,容易發(fā)現(xiàn)數(shù)列bn是以-3為首項(xiàng),-3為公比的等比數(shù)列,記數(shù)列an的所有項(xiàng)的和為S,數(shù)列bn的所有項(xiàng)的和為T(mén),則()A.S>TB.S=TC.S<TD.S與T的大小關(guān)系不確定【解析】選A.因?yàn)閍n=(2n-1)(-1)n+1,所以an+an+1=(2n-1)(-1)n+1+(2n+1)(-1)n+2=(-1)n+2-(2n-1)+(2n+1)=2(-1)n,所以S=a1+(a2+a3)+(a4+a5)+(a728+a729)=1+2+2+2364個(gè)2=729,因?yàn)閍728=-1 455,a729=1 457,又因?yàn)閿?shù)列bn是以-3為首項(xiàng),-3為公比的等比數(shù)列,所以b6=(-3)6=729,所以數(shù)列bn共有6項(xiàng),所以所有項(xiàng)的和為T(mén)=-3(1-(-3)6)1-(-3)=546,所以S>T.3.已知數(shù)列an的前n項(xiàng)和為Sn,若a1為函數(shù)f(x)=3sin x+cos x(xR)的最大值,且滿足an-anSn+1=a12-anSn,則數(shù)列an的前2 018項(xiàng)之積A2 018=()A.1B.12C.-1D.2【解析】選A.因?yàn)閍1為函數(shù)f(x)=3sin x+cos x(xR)的最大值,所以a1=2,因?yàn)閍n-anSn+1=a12-anSn,所以(Sn+1-Sn)an=an-1,所以an+1=an-1an,所以a2=12,a3=-1,a4=2所以數(shù)列an是周期數(shù)列,周期為3,所以an的前2018項(xiàng)之積A2018=a1a2a3a2 018=(-1)6721=1.4.(2018 河南省六市聯(lián)考)已知數(shù)列an滿足: an+1+(-1)n+1an=2,則其前100項(xiàng)的和為 ()A.250B.200C.150D.100【解析】選D.因?yàn)閍n+1+(-1)n+1an=2,所以a1+a2=2,a3+a4=2,a5+a6=2,a99+a100=2,所以其前100項(xiàng)和為250=100.5.已知數(shù)列an滿足a1=1, an+1-an=4n-2(nN*),則使an163的正整數(shù)n的最小值為()A.8B.10C.12D.14【解析】選B.由題意得an+1-an=4n-2,則當(dāng)n2時(shí),a2-a1=2,a3-a2=6,an-an-1=4n-6,這n-1個(gè)式子相加,就有an-a1=(n-1)(2+4n-6)2=2(n-1)2,即an=2(n-1)2+1=2n2-4n+3,當(dāng)n=1時(shí),a1=1也滿足上式,所以an=2n2-4n+3,由an163得2n2-4n+3163,即n2-2n-800,解得n10或n-8,所以n10,即使an163的正整數(shù)n的最小值為10.二、填空題(每小題5分,共15分)6.已知數(shù)列an的通項(xiàng)公式為an=1(n+1)n+nn+1(nN*),記數(shù)列an的前n項(xiàng)和Sn,則在S1,S2,S2 017中,有_個(gè)有理數(shù).【解析】依題意,an=1(n+1)n+nn+1=1nn+1(n+1+n)=n+1-nnn+1=1n-1n+1,故Sn=a1+a2+an=1-1n+1,因?yàn)?4<2 018<45,故n+1=22,32,442,故有43個(gè)有理數(shù).答案:437.(2018湖北省聯(lián)考試題)“斐波那契數(shù)列”由十三世紀(jì)意大利數(shù)學(xué)家列昂納多斐波那契發(fā)現(xiàn),因?yàn)殪巢瞧跻酝米臃敝碁槔佣?故又稱該數(shù)列為“兔子數(shù)列”.斐波那契數(shù)列an滿足:a1=1,a2=1,an=an-1+an-2(n3,nN*),記其前n項(xiàng)和為Sn,設(shè)a2 018=t(t為常數(shù)),則S2 016+S2 015-S2 014-S2 013=_(用t表示).【解析】S2 016+S2 015-S2 014-S2 013=a2 016+a2 015+a2 015+a2 014=a2 017+a2 016=a2 018=t.答案:t8.有下列命題:等比數(shù)列an中,前n項(xiàng)和為Sn,公比為q,則Sn,S2n-Sn,S3n-S2n,仍然是等比數(shù)列,其公比為qn;若數(shù)列an是等差數(shù)列,前n項(xiàng)和為Sn,則數(shù)列Snn是等差數(shù)列;若數(shù)列an是正項(xiàng)數(shù)列,且a1+a2+an=n2+3n(nN*),則a12+a23+ann+1=2n2+6n;若數(shù)列an的前n項(xiàng)和為Sn,Sn=20n-19,則數(shù)列an是等比數(shù)列.其中正確命題的序號(hào)是_(填序號(hào)).【解析】錯(cuò),q=-1,n=2,S4-S2=0,不符合等比數(shù)列. 因?yàn)閿?shù)列an是等差數(shù)列,前n項(xiàng)和為Sn,所以Sn=na1+n(n-1)2d,所以Snn=a1+(n-1)d2,所以數(shù)列Snn是以a1為首項(xiàng),以d2為公差的等差數(shù)列.a1+a2+an=n2+3n中n用n-1代替得a1+a2+an-1=(n-1)2+3(n-1),兩式作差得an=2n+2(n2),an=4(n+1)2,a1=16,符合.ann+1=4(n+1),所以a12+a23+ann+1=2n2+6n.當(dāng)n=1時(shí), a1=S1=201-19=20-18,當(dāng)n2時(shí),an=Sn-Sn-1=20n-19-20n-20=1920n-20,所以an=20-18,n=11920n-20,n2,所以數(shù)列an不是等比數(shù)列答案:三、解答題(每小題10分,共30分)9.(2018亳州市一模)已知數(shù)列an的前n項(xiàng)和Sn滿足2Sn=3an-,其中是不為零的常數(shù),nN*.(1)求an的通項(xiàng)公式.(2)若=3,記bn=2log3an,求數(shù)列bnbn+2的前n項(xiàng)和Tn.【解析】(1)由已知2Sn=3an-可得:2Sn+1=3an+1-兩式相減得:2an+1=3an+1-3an,即an+1=3an因?yàn)?S1=3a1-,所以a1=0,所以an0,所以an+1an=3,所以an是首項(xiàng)為,公比為3的等比數(shù)列,從而an=3n-1.(2)因?yàn)?3,所以an=3n,從而bn=2n,所以bnbn+2=4n(n+2)=21n-1n+2,所以Tn=21-13+12-14+13-15+1n-1n+2=21+12-1n+1-1n+2=3-2n+1-2n+2.10.設(shè)數(shù)列an的前n項(xiàng)和為Sn,a1=1,且對(duì)任意正整數(shù)n,點(diǎn)(an+1,Sn)都在直線2x+y-2=0上.(1)求數(shù)列an的通項(xiàng)公式.(2)若bn=nan2,數(shù)列bn的前n項(xiàng)和為T(mén)n,求證:Tn<169.【解析】(1)因?yàn)辄c(diǎn)(an+1,Sn)在直線2x+y-2=0上,所以2an+1+Sn-2=0,當(dāng)n>1時(shí),2an+Sn-1-2=0,兩式相減得2an+1-2an+Sn-Sn-1=0,即2an+1-2an+an=0,an+1=12an,又當(dāng)n=1時(shí),2a2+S1-2=2a2+a1-2=0,a2=12=12a1,所以an是首項(xiàng)a1=1,公比q=12的等比數(shù)列,數(shù)列an的通項(xiàng)公式為an=12n-1.(2)由(1)知,bn=nan2=n4n-1,則Tn=1+24+342+n4n-1,14Tn=14+242+n-14n-1+n4n,兩式相減得34Tn=1+14+142+14n-1-n4n=1-14n1-14-n4n=431-14n-n4n,所以Tn=169-16+12n94n<169.11.設(shè)等差數(shù)列an的公差為d,點(diǎn)(an,bn)在函數(shù)f(x)=2x的圖象上(nN*).(1)若a1=-2,點(diǎn)(a8,4b7)在函數(shù)f(x)的圖象上,求數(shù)列an的前n項(xiàng)和Sn.(2)若a1=1,函數(shù)f(x)的圖象在點(diǎn)(a2,b2)處的切線在x軸上的截距為2-1ln2,求數(shù)列anbn的前n項(xiàng)和Tn.【解析】(1)點(diǎn)(an,bn)在函數(shù)f(x)=2x的圖象上,所以bn=2an,又等差數(shù)列an的公差為d,所以bn+1bn=2an+12an=2an+1-an=2d,因?yàn)辄c(diǎn)(a8,4b7)在函數(shù)f(x)的圖象上,所以4b7=2a8=b8,所以2d=b8b7=4d=2.又a1=-2,所以 Sn=na1+n(n-1)2d=-2n+n2-n=n2-3n.(2)由f(x)=2xf(x)=2xln 2,函數(shù)f(x)的圖象在點(diǎn)(a2,b2)處的切線方程為y-b2=(2a2ln 2)(x-a2),所以切線在x軸上的截距為a2-1ln2,從而a2-1ln2=2-1ln2,故a2=2,從而an=n,bn=2n,anbn=n2n,Tn=12+222+322+n2n,Tn=122+223+324+n2n+1,所以12Tn=12+122+123+124+12n-n2n+1=1-12n-n2n+1=1-n+22n+1,故Tn=2-n+22n.(20分鐘20分)1.(10分)設(shè)數(shù)列an滿足a1=0且11-an+1-11-an=1.(1)求an的通項(xiàng)公式.(2)設(shè)bn=1-an+1n,記Sn=k=1nbk,證明:Sn<1.【解析】(1)由題設(shè)11-an+1-11-an=1,即11-an是公差為1的等差數(shù)列.又11-a1=1,故11-an=n.所以an=1-1n.(2)由(1)得bn=1-an+1n=n+1-nn+1n=1n-1n+1,Sn=k=1nbk=k=1n1k-1k+1=1-1n+1<1.2.(10分)已知數(shù)列an的前n項(xiàng)和是Sn,且Sn+13an=1(nN*).(1)求數(shù)列an的通項(xiàng)公式.(2)設(shè)bn=log4(1-Sn+1)(nN*),Tn=1b1b2+1b2b3+1bnbn+1,求使Tn1 0072 016成立的最小的正整數(shù)n的值.【解析】(1)當(dāng)n=1時(shí),a1=S1,由S1+13a1=1a1=34,當(dāng)n2時(shí),Sn+13an=1,Sn-1+13an-1=1,-,得an+13an-13an-1=0,即an=14an-1,所以an是以34為首項(xiàng),14為公比的等比數(shù)列.故an=3414n-1=314n(nN*).(2)由(1)知1-Sn+1=13an+1=14n+1,bn=log4(1-Sn+1)=log414n+1=-(n+1),1bnbn+1=1(n+1)(n+2)=1n+1-1n+2,Tn=1b1b2+1b2b3+1bnbn+1=12-13+13-14+1n+1-1n+2=12-1n+2,12-1n+21 0072 016n2 014,故使Tn1 0072 016成立的最小的正整數(shù)n的值為2 014.