题型练4大题专项(二)数列的通项、求和问题1.(2019湖北4月调研,17)已知数列{an}满足a2-a1=1,其前n项和为Sn,当n≥2时,Sn-1-1,Sn,Sn+1成等差数列.(1)求证{an}为等差数列;(2)若Sn=0,Sn+1=4,求n.2.(2019吉林实验中学检测,17)已知等差数列{an}满足a4=7,2a3+a5=19.(1)求an;(2)设{bn-an}是首项为2,公比为2的等比数列,求数列{bn}的通项公式及前n项和Tn.3.设{an}是等差数列,且a1=ln2,a2+a3=5ln2.(1)求{an}的通项公式.(2)求ea1+ea2+…+ean.4.已知等差数列{an}的前n项和为Sn,公比为q的等比数列{bn}的首项是12,且a1+2q=3,a2+4b2=6,S5=40.(1)求数列{an},{bn}的通项公式an,bn;(2)求数列{1anan+1+1bnbn+1}的前n项和Tn.5.已知函数f(x)=7x+5x+1,数列{an}满足:2an+1-2an+an+1an=0,且anan+1≠0.在数列{bn}中,b1=f(0),且bn=f(an-1).(1)求证:数列{1an}是等差数列;(2)求数列{|bn|}的前n项和Tn.6.已知等比数列{an}的公比q>1,且a3+a4+a5=28,a4+2是a3,a5的等差中项.数列{bn}满足b1=1,数列{(bn+1-bn)an}的前n项和为2n2+n.(1)求q的值;(2)求数列{bn}的通项公式.题型练4大题专项(二)数列的通项、求和问题1.(1)证明当n≥2时,由Sn-1-1,Sn,Sn+1成等差数列,可知2Sn=Sn-1-1+Sn+1,即Sn-Sn-1=-1+Sn+1-Sn,即an=-1+an+1(n≥2),则an+1-an=1(n≥2),又a2-a1=1,故{an}是公差为1的等差数列.(2)解由(1)知等差数列{an}的公差为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)由题意得{a1+3d=7,2(a1+2d)+a1+4d=19,解得{a1=1,d=2.∴an=1+2(n-1)=2n-1.(2)由题意可知bn-an=2n,∴bn=2n+2n-1,∴Tn=(2+22+…+2n)+[1+3+…+(2n-1)],∴Tn=2n+1+n2-2.3.解(1)设等差数列{an}的公差为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为首项,2为公比的等比数列.∴ea1+ea2+…+ean=2+22+…+2n=2n+1-2.∴ea1+ea2+…+ean=2n+1-2.4.解(1)设{an}公差为d,由题意得{a1+2d=8,a1+2q=3,a1+d+2q=6,解得{a1=2,d=3,q=12,故an=3n-1,bn=(12)n.(2)∵1anan+1+1bnbn+1=13(1an-1an+1)+1bnbn+1=13(1an-1an+1)+22n+1,∴Tn=13[(12-15)+(15-18)+…+(13n-1-13n+2)]+8(1-4n)1-4=13(12-13n+2)+13(22n+3-8)=13(22n+3-13n+2)−52.5.(1)证明∵2an+1-2an+an+1an=0,∴1an+1−1an=12,故数列{1an}是以12为公差的等差数列.(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.当n≤6时,Tn=n2(5+6-n)=n(11-n)2;当n≥7时,Tn=15+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的等差中项,得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.由a3+a5=20,得8(q+1q)=20,解得q=2或q=12,因为q>1,所以q=2.(2)设cn=(bn+1-bn)an,数列{cn}前n项和为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)·(12)n-1.故bn-bn-1=(4n-5)·(12)n-2,n≥2,bn-b1=(bn-bn-1)+(bn-1-bn-2)+…+(b3-b2)+(b2-b1)=(4n-5)·(12)n-2+(4n-9)·(12)n-3+…+7·12+3.设Tn=3+7·12+11·(12)2+…+(4n-5)·(12)n-2,n≥2,12Tn=3·12+7·(12)2+…+(4n-9)·(12)n-2+(4n-5)·(12)n-1,所以12Tn=3+4·12+4·(12)2+…+4·(12)n-2-(4n-5)·(12)n-1,因此Tn=14-(4n+3)·(12)n-2,n≥2,又b1=1,所以bn=15-(4n+3)·(12)n-2.
