发布时间 : 星期四 文章(全国通用版)2019高考数学二轮复习中档大题规范练(二)数列(理)更新完毕开始阅读
(二)
数 列
1.(2018·三明质检)已知正项数列{an}的前n项和为Sn,a1=1,且(t+1)Sn=an+3an+2(t∈R).
(1)求数列{an}的通项公式;
?1???的前n项和Tn. (2)若数列{bn}满足b1=1,bn+1-bn=an+1,求数列
?2bn+7n?
2
解 (1)因为a1=1,且(t+1)Sn=an+3an+2, 所以(t+1)S1=a1+3a1+2,所以t=5. 所以6Sn=an+3an+2.①
当n≥2时,有6Sn-1=an-1+3an-1+2,② ①-②得6an=an+3an-an-1-3an-1, 所以(an+an-1)(an-an-1-3)=0, 因为an>0,所以an-an-1=3, 又因为a1=1,
所以{an}是首项a1=1,公差d=3的等差数列, 所以an=3n-2(n∈N). (2)因为bn+1-bn=an+1,b1=1, 所以bn-bn-1=an(n≥2,n∈N), 所以当n≥2时,
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*
2
22
2
2
2
bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1
3n-n=an+an-1+…+a2+b1=.
2
3n-n*
又b1=1也适合上式,所以bn=(n∈N).
211
所以=2
2bn+7n3n-n+7n1?111?1
=·=·?-?, 3n?n+2?6?nn+2?
11?1?111
所以Tn=·?1-+-+…+-
324nn+2?6??11?1?3
-=·?-?,
6?2n+1n+2?3n+5n=. 12?n+1??n+2?
2
2
2
2.(2018·葫芦岛模拟)设等差数列{an}的前n项和为Sn,且S3,,S4成等差数列,a5=3a2
2+2a1-2.
(1)求数列{an}的通项公式; (2)设bn=2
n-1
S5
,求数列??的前n项和Tn.
?bn?
?an?
解 (1)设等差数列{an}的首项为a1,公差为d, 由S3,,S4成等差数列,
2可知S3+S4=S5,得2a1-d=0,① 由a5=3a2+2a1-2,② 得4a1-d-2=0,
由①②,解得a1=1,d=2, 因此,an=2n-1(n∈N).
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S5
an?1?n-1
(2)令cn==(2n-1)??,
bn?2?
则Tn=c1+c2+…+cn,
1?1?2?1?n-1
∴Tn=1·1+3·+5·??+…+(2n-1)·??,③
2?2??2?1?21?31?n11???Tn=1·+3·??+5·??+…+(2n-1)·??,④
22?2??2??2?1?1?1?2?1?n-1??1?n③-④,得Tn=1+2?+??+…+???-(2n-1)·??
2?2???2??2?2?2n+3??1?n-1??1?n=1+2?1-??? -(2n-1)·??= 3-n,
2??2???2?2n+3*
∴Tn=6-n-1(n∈N).
2
3.(2018·厦门质检)已知等差数列{an}满足(n+1)an=2n+n+k,k∈R. (1)求数列{an}的通项公式; (2)设bn=
4n2
2
anan+1
,求数列{bn}的前n项和Sn.
2
解 (1)方法一 由(n+1)an=2n+n+k, 令n=1,2,3,
3+k10+k21+k得到a1=,a2=,a3=,
234∵{an}是等差数列,∴2a2=a1+a3,
即
20+2k3+k21+k=+, 324
解得k=-1.
由于(n+1)an=2n+n-1=(2n-1)(n+1), 又∵n+1≠0,∴an=2n-1(n∈N). 方法二 ∵{an}是等差数列,设公差为d, 则an=a1+d(n-1)=dn+(a1-d), ∴(n+1)an=(n+1)(dn+a1-d) =dn+a1n+a1-d,
∴dn+a1n+a1-d=2n+n+k对于?n∈N均成立,
2
2
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2
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2
d=2,??
则?a1=1,??a1-d=k,
4n2
解得k=-1,∴an=2n-1(n∈N).
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4n(2)由bn==
anan+1?2n-1??2n+1?4n1=2=1+2 4n-14n-1
1?11?1
-=1+=??+1,
?2n-1??2n+1?2?2n-12n+1?得Sn=b1+b2+b3+…+bn
1?1?1?1?11?1?11?1?1-=?1-?+1+?-?+1+?-?+1+…+??+1 3?2?2?35?2?57?2?2n-12n+1?11?1?11111
-=?1-+-+-+…++n
335572n-12n+1?2??1?1?
=?1-+n
2n+1?2??
2n2n+2n*
=+n=(n∈N). 2n+12n+1
2
2
4.(2018·天津河东区模拟)已知等比数列{an}满足条件a2+a4=3(a1+a3),a2n=3an,n∈N. (1)求数列{an}的通项公式;
(2)数列{bn}满足++…+=n,n∈N,求{bn}的前n项和Tn. 解 (1)设{an}的通项公式为an=a1q由已知a2+a4=3(a1+a3),
得a1q+a1q=3(a1+a1q),所以q=3. 又由已知a2n=3an,
2
3
2
2*
b1b2
a1a2bnan2*
n-1
(n∈N),
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