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1、題型練4　大題專(zhuān)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(mǎn)足a2-a1=1,其前n項(xiàng)和為Sn,當(dāng)n≥2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(mǎn)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

2、)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(mǎn)足:2an+1-2an+an+1an=0,且anan+1≠0.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

3、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(mǎn)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(zhuān)項(xiàng)(二) 數(shù)列的通項(xiàng)、求和問(wèn)題 1.(1)證明當(dāng)n≥2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(n≥2),則an+1-an=1(n≥2), 又a2-a1=1,故{an}是公差為1的等差數(shù)列. (

4、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,

5、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

6、+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+2?52. 5.(1)證明∵2an+1-2an+an+1an=0,∴1an+1?1an=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)n≤6時(shí),Tn=n2(5+6-n)=n(11-n)2; 當(dāng)n≥7時(shí),Tn=15

7、+n-62(1+n-6)=n2-11n+602.故Tn=n(11-n)2,n≤6,n2-11n+602,n≥7. 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,n≥2,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,n≥2, 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,n≥2, 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,n≥2, 又b1=1,所以bn=15-(4n+3)·12n-2.