2021屆高考數(shù)學二輪復(fù)習題型練4大題專項二數(shù)列的通項求和問題文含解析

1、題型練4大題專項(二)數(shù)列的通項、求和問題題型練第56頁1.已知數(shù)列an滿足a2-a1=1,其前n項和為sn,當n2時,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.(1)證明當n2時,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-

2、1)2=0,即a1+n-12=0,解得n=7.2.已知等差數(shù)列an滿足a4=7,2a3+a5=19.(1)求an;(2)設(shè)bn-an是首項為2,公比為2的等比數(shù)列,求數(shù)列bn的通項公式及前n項和tn.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.設(shè)an是等差數(shù)列,且a1=ln 2,a2+a3=5ln 2.(1)求an的通項公式.(2)求ea1+ea2+ean.解(1)設(shè)等差數(shù)列an的公

3、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,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.已知等差數(shù)列an的前n項和為sn,公比為q的等比數(shù)列bn的首項是12,且a1+2q=3,a2+4b2=6,s5=40.(1)求數(shù)列an,bn的通項公式an,bn;(2)求數(shù)列1anan+1+1bnbn+1的前n項和tn.解(1)設(shè)an公差為d,由題意得a1+2d=8,a1+2

4、q=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+2-52.5.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項和tn.(

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n6時,tn=n2(5+6-n)=n(11-n)2;當n7時,tn=15+n-62(1+n-6)=n2-11n+602.故tn=n(11-n)2,n6,n2-11n+602,n7.6.已知等比數(shù)列an的公比q1,且a3+a4+a5=28,a4+2是a3,a5的等差中項.數(shù)列bn滿足b1=1,數(shù)列(bn+

6、1-bn)an的前n項和為2n2+n.(1)求q的值;(2)求數(shù)列bn的通項公式.解(1)由a4+2是a3,a5的等差中項,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,因為q1,所以q=2.(2)設(shè)cn=(bn+1-bn)an,數(shù)列cn前n項和為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+712+3.設(shè)tn=3+712+11122+(4n-5)12n-2,n2,12tn=312+7122+(4n-9)12n-2