6.(2019西藏山南地区第二高级中学高三上学期期中模拟)已知等差数列{an}的前n项和为Sn,且

S9=90,S15=240.

(1)求数列{an}的通项公式和前n项和Sn;

(2)设{bn-(-1)an}是等比数列,且b2=7,b5=71,求数列{bn}的前n项和Tn.

7.(2019山东烟台高三5月适应性练习)已知数列{an}前n项和Sn满足Sn=2an-2(n∈N),{bn}是等差数列,且a3=b4-2b1,b6=a4. (1)求{an}和{bn}的通项公式;

(2)求数列{(-1)??2??}的前2n项和T2n.

n*n 5

8.设{an}是等差数列,其前n项和为Sn(n∈N);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6. (1)求Sn和Tn;

(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.

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参考答案

专题突破练13 等差、等比

数列的综合问题

1.解(1)∵a1

n+1=3

an+2,a1=4,

∴a1

n+1-3=3(an-3).

故{a3}是首项为1,公比为1

n-3的等比数列.

∴a1

n=3+1n-3

.

(2)a1n-1

n=3+3

,

故T0

n-1

1-(1

n=3n+13) ??3

+11

3

+…+13

=3n+1-1=3n+31n3

21-3

.

2.(1)证明当n≥2时,由Sn-1-1,Sn,Sn+1成等差数列,得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)知数列{an}的公差为1. 由Sn=0,Sn+1=4,得an+1=4,即a1+n=4,

由S??-1)

??-1n=0,得na1+??(2

=0,即a1+2

=0,

7

联立解得n=7.

4×3

3.(1)解由题意得{4??1+2??=16,

(????)2

1+=??1(??1+4??),由于d≠0,解得{

??1=1,

??=2.

∴an=1+(n-1)×2=2n-1.

(2)证明由(1)知S??(??-1)

2

n=n×1+2

×2=n,

∴△ABC的面积S=12(Sn+Sn+2)×2-11

2(Sn+Sn+1)×1-2(Sn+1+Sn+2)×1

=1

2(Sn+Sn+2-2Sn+1)

=1

2[n2+(n+2)2-2(n+1)2]=1.

4.解(1)∵3S1,2S2,S3成等差数列,∴4S2=3S1+S3,

∴4(a1+a2)=3a1+(a1+a2+a3),

即a3=3a2,∴公比q=3,

∴a1nn=a1qn-=3.

(2)由(1)知,bnn=log3an=log33=n,

∵b2n-1b2n-b2nb2n+1=(2n-1)·2n-2n(2n+1)=-4n,

∴T)+(b??(??+1)

n=(b1b2-b2b33b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)=-4(1+2+…+n)=-4×2

=-2n2-2n.5.解(1)由题意得an-1n+1

1+2a2+…+2an=n·2, 所以a2

1=1×2=4,a3

1+2a2=2×2,得a2=6.

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