高考數(shù)學(xué)（理）二輪專題對(duì)點(diǎn)練13　等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和

1、高考數(shù)學(xué)（理）二輪專題對(duì)點(diǎn)練13等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和專題對(duì)點(diǎn)練13等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和專題對(duì)點(diǎn)練第17頁1.Sn為數(shù)列an的前n項(xiàng)和.已知an0,an2+2an=4Sn+3.(1)求an的通項(xiàng)公式;(2)設(shè)bn=1anan+1,數(shù)列bn的前n項(xiàng)和為Tn,求Tn.解 (1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.兩式相減可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由于an0,因此an+1-an=2.又a12+2a1=4a1+3,解得a1=3(a

2、1=-1舍去).所以an是首項(xiàng)為3,公差為2的等差數(shù)列,故an=2n+1.(2)由an=2n+1可知bn=1anan+1=1212n+1-12n+3.Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).2.已知數(shù)列an是等差數(shù)列,前n項(xiàng)和為Sn,若a1=9,S3=21.(1)求數(shù)列an的通項(xiàng)公式;(2)若a5,a8,Sk成等比數(shù)列,求k的值.解 (1)設(shè)等差數(shù)列an的公差為d,a1=9,S3=21,S3=39+322d=21,解得d=-2,an=9+(n-1)(-2)=-2n+11.(2)a5,a8,Sk成等比數(shù)列,a82=a5Sk,即(-28+11)

3、2=(-25+11)9k+k(k-1)2(-2),解得k=5.3.(2017河北衡水中學(xué)三調(diào),理17)已知數(shù)列an的前n項(xiàng)和為Sn,a10,常數(shù)0,且a1an=S1+Sn對(duì)一切正整數(shù)n都成立.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)a10,=100,當(dāng)n為何值時(shí),數(shù)列l(wèi)g1an的前n項(xiàng)和最大?解 (1)令n=1,得a12=2S1=2a1,即a1(a1-2)=0.因?yàn)閍10,所以a1=2.當(dāng)n2時(shí),2an=2+Sn,2an-1=2+Sn-1,兩式相減,得2an-2an-1=an(n2),所以an=2an-1(n2),從而數(shù)列an為等比數(shù)列,所以an=a12n-1=2n.(2)當(dāng)a10,=100時(shí),由

4、(1)知,an=2n100,設(shè)bn=lg1an=lg1002n=lg 100-lg 2n=2-nlg 2,所以數(shù)列bn是單調(diào)遞減的等差數(shù)列,公差為-lg 2,所以b1b2b6=lg10026=lg10064lg 1=0,當(dāng)n7時(shí),bnb7=lg100271,且a1+a3=20,a2=8.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=nan,Sn是數(shù)列bn的前n項(xiàng)和,求Sn.解 (1)等比數(shù)列an的公比q1,且a1+a3=20,a2=8,a1+a1q2=20,a1q=8,2q2-5q+2=0,解得q=2,a1=4.an=2n+1.(2)bn=nan=n2n+1,Sn=122+223+n2n+1,12

5、Sn=123+224+n-12n+1+n2n+2.12Sn=122+123+12n+1-n2n+2=141-12n1-12-n2n+2=12-2+n2n+2.Sn=1-2+n2n+1.6.(2017安徽安慶二模,理17)在數(shù)列an中,a1=2,a2=4,設(shè)Sn為數(shù)列an的前n項(xiàng)和,對(duì)于任意的n1,nN*,Sn+1+Sn-1=2(Sn+1).(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=n2an,求bn的前n項(xiàng)和Tn.解 (1)對(duì)于任意的n1,nN*,Sn+1+Sn-1=2(Sn+1),Sn+2+Sn=2(Sn+1+1),兩式相減可得an+2+an=2an+1.(*)當(dāng)n=2時(shí),S3+S1=2(S2

6、+1),即2a1+a2+a3=2(a1+a2+1),解得a3=6.當(dāng)n=1時(shí)(*)也滿足.數(shù)列an是等差數(shù)列,公差為2,an=2+2(n-1)=2n.(2)bn=n2an=n22n=n4n,Tn=14+242+343+n4n,14Tn=142+243+n-14n+n4n+1,34Tn=14+142+14n-n4n+1=141-14n1-14-n4n+1,Tn=49-4+3n94n.7.(2017山東,理19)已知xn是各項(xiàng)均為正數(shù)的等比數(shù)列,且x1+x2=3,x3-x2=2.(1)求數(shù)列xn的通項(xiàng)公式;(2)如圖,在平面直角坐標(biāo)系xOy中,依次連接點(diǎn)P1(x1,1),P2(x2,2)Pn+1(

7、xn+1,n+1)得到折線P1P2Pn+1,求由該折線與直線y=0,x=x1,x=xn+1所圍成的區(qū)域的面積Tn.解 (1)設(shè)數(shù)列xn的公比為q,由已知q0.由題意得x1+x1q=3,x1q2-x1q=2.所以3q2-5q-2=0.因?yàn)閝0,所以q=2,x1=1,因此數(shù)列xn的通項(xiàng)公式為xn=2n-1.(2)過P1,P2,Pn+1向x軸作垂線,垂足分別為Q1,Q2,Qn+1.由(1)得xn+1-xn=2n-2n-1=2n-1,記梯形PnPn+1Qn+1Qn的面積為bn,由題意bn=(n+n+1)22n-1=(2n+1)2n-2,所以Tn=b1+b2+bn=32-1+520+721+(2n-1)

8、2n-3+(2n+1)2n-2.又2Tn=320+521+722+(2n-1)2n-2+(2n+1)2n-1,-得-Tn=32-1+(2+22+2n-1)-(2n+1)2n-1=32+2(1-2n-1)1-2-(2n+1)2n-1.所以Tn=(2n-1)2n+12.8.(2017山東濰坊一模,理19)已知數(shù)列an是等差數(shù)列,其前n項(xiàng)和為Sn,數(shù)列bn是公比大于0的等比數(shù)列,且b1=-2a1=2,a3+b2=-1,S3+2b3=7.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)令cn=2,n為奇數(shù),-2anbn,n為偶數(shù),求數(shù)列cn的前n項(xiàng)和Tn.解 (1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比

9、為q,q0,b1=-2a1=2,a3+b2=-1,S3+2b3=7,a1=-1,b1=2,-1+2d+2q=-1,3(-1)+3d+22q2=7,解得d=-2,q=2.an=-1-2(n-1)=1-2n,bn=2n.(2)cn=2,n為奇數(shù),2n-12n-1,n為偶數(shù).當(dāng)n=2k(kN*)時(shí),數(shù)列cn的前n項(xiàng)和Tn=T2k=(c1+c3+c2k-1)+(c2+c4+c2k)=2k+32+723+4k-122k-1,令A(yù)k=32+723+4k-122k-1,14Ak=323+725+4k-522k-1+4k-122k+1,34Ak=32+4123+125+122k-1-4k-122k+1=32+4181-14k-11-14-4k-122k+1,Ak=269-12k+13922k-1.Tn=T2k=2k+269-12k+13922k-1.當(dāng)n=