題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問(wèn)題1.(2019湖北4月調(diào)研,17)已知數(shù)列an滿足a2-a1=1,其前n項(xiàng)和為Sn,當(dāng)n2時(shí),Sn-1-1,Sn,Sn+1成等差數(shù)列.(1)求證an為等差數(shù)列;(2)若Sn=0,Sn+1=4,求n.2.(2019吉林實(shí)驗(yàn)中學(xué)檢測(cè),17)已知等差數(shù)列an滿足a4=7,2a3+a5=19.(1)求an;(2)設(shè)bn-an是首項(xiàng)為2,公比為2的等比數(shù)列,求數(shù)列bn的通項(xiàng)公式及前n項(xiàng)和Tn.3.設(shè)an是等差數(shù)列,且a1=ln 2,a2+a3=5ln 2.(1)求an的通項(xiàng)公式.(2)求ea1+ea2+ean.4.已知等差數(shù)列an的前n項(xiàng)和為Sn,公比為q的等比數(shù)列bn的首項(xiàng)是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求數(shù)列an,bn的通項(xiàng)公式an,bn;(2)求數(shù)列1anan+1+1bnbn+1的前n項(xiàng)和Tn.5.已知函數(shù)f(x)=7x+5x+1,數(shù)列an滿足:2an+1-2an+an+1an=0,且anan+10.在數(shù)列bn中,b1=f(0),且bn=f(an-1).(1)求證:數(shù)列1an是等差數(shù)列;(2)求數(shù)列|bn|的前n項(xiàng)和Tn.6.已知等比數(shù)列an的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列bn滿足b1=1,數(shù)列(bn+1-bn)an的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列bn的通項(xiàng)公式.題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問(wèn)題1.(1)證明當(dāng)n2時(shí),由Sn-1-1,Sn,Sn+1成等差數(shù)列,可知2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-1+Sn+1-Sn,即an=-1+an+1(n2),則an+1-an=1(n2),又a2-a1=1,故an是公差為1的等差數(shù)列.(2)解由(1)知等差數(shù)列an的公差為1.由Sn=0,Sn+1=4,得an+1=4,即a1+n=4.由Sn=0,得na1+n(n-1)2=0,即a1+n-12=0,解得n=7.2.解(1)由題意得a1+3d=7,2(a1+2d)+a1+4d=19,解得a1=1,d=2.an=1+2(n-1)=2n-1.(2)由題意可知bn-an=2n,bn=2n+2n-1,Tn=(2+22+2n)+1+3+(2n-1),Tn=2n+1+n2-2.3.解(1)設(shè)等差數(shù)列an的公差為d,a2+a3=5ln2.2a1+3d=5ln2,又a1=ln2,d=ln2.an=a1+(n-1)d=nln2.(2)由(1)知an=nln2.ean=enln2=eln2n=2n,ean是以2為首項(xiàng),2為公比的等比數(shù)列.ea1+ea2+ean=2+22+2n=2n+1-2.ea1+ea2+ean=2n+1-2.4.解(1)設(shè)an公差為d,由題意得a1+2d=8,a1+2q=3,a1+d+2q=6,解得a1=2,d=3,q=12,故an=3n-1,bn=12n.(2)1anan+1+1bnbn+1=131an-1an+1+1bnbn+1=131an-1an+1+22n+1,Tn=1312-15+15-18+13n-1-13n+2+8(1-4n)1-4=1312-13n+2+13(22n+3-8)=1322n+3-13n+252.5.(1)證明2an+1-2an+an+1an=0,1an+11an=12,故數(shù)列1an是以12為公差的等差數(shù)列.(2)解b1=f(0)=5,7(a1-1)+5a1-1+1=5,7a1-2=5a1,a1=1,1an=1+(n-1)·12,an=2n+1,bn=7an-2an=7-(n+1)=6-n.當(dāng)n6時(shí),Tn=n2(5+6-n)=n(11-n)2;當(dāng)n7時(shí),Tn=15+n-62(1+n-6)=n2-11n+602.故Tn=n(11-n)2,n6,n2-11n+602,n7.6.解(1)由a4+2是a3,a5的等差中項(xiàng),得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8q+1q=20,解得q=2或q=12,因?yàn)閝>1,所以q=2.(2)設(shè)cn=(bn+1-bn)an,數(shù)列cn前n項(xiàng)和為Sn,由cn=S1,n=1,Sn-Sn-1,n2,解得cn=4n-1.由(1)可知an=2n-1,所以bn+1-bn=(4n-1)·12n-1.故bn-bn-1=(4n-5)·12n-2,n2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+(b3-b2)+(b2-b1)=(4n-5)·12n-2+(4n-9)·12n-3+7·12+3.設(shè)Tn=3+7·12+11·122+(4n-5)·12n-2,n2,12Tn=3·12+7·122+(4n-9)·12n-2+(4n-5)·12n-1,所以12Tn=3+4·12+4·122+4·12n-2-(4n-5)·12n-1,因此Tn=14-(4n+3)·12n-2,n2,又b1=1,所以bn=15-(4n+3)·12n-2.