题目
题型:不详难度:来源:
(1)求数列{an}的通项an;
(2)若数列{bn}的满足bn=log2(an+2),Tn为数列{
bn |
an+2 |
1 |
2 |
答案
①-②,得an=2an-2an-1-2,即an=2an-1+2,
∴an+2=2(an-1+2),∴
an+2 |
an-1+2 |
当n=1时,S1=2a1-2,则a1=2.
∴{an+2}是以a1+2=4为首项,2为公比的等比数列,
∴an+2=4•2n-1,∴an=2n+1-2;
(2)证明:bn=log2(an+2)=log22n+1=n+1,∴
bn |
an+2 |
n+1 |
2n+1 |
则Tn=
2 |
22 |
3 |
23 |
n+1 |
2n+1 |
1 |
2 |
2 |
23 |
3 |
24 |
n |
2n+1 |
n+1 |
2n+2 |
③-④,得
1 |
2 |
2 |
22 |
1 |
23 |
1 |
24 |
1 |
2n+1 |
n+1 |
2n+2 |
1 |
2 |
| ||||
1-
|
n+1 |
2n+2 |
3 |
4 |
n+3 |
2n+2 |
∴Tn=
3 |
2 |
n+3 |
2n+1 |
当n≥2时,Tn-Tn-1=-
n+3 |
2n+1 |
n+2 |
2n |
n+1 |
2n+1 |
∴{Tn}为递增数列,∴Tn≥T1=
1 |
2 |
核心考点
试题【已知数列{an}的前n项和Sn,满足:Sn=2an-2n(n∈N*).(1)求数列{an}的通项an;(2)若数列{bn}的满足bn=log2(an+2),Tn】;主要考察你对数列综合等知识点的理解。[详细]
举一反三
Sn |
(1)求a2,a3;
(2)求数列{an}的通项公式an;
(3)设bn=
1 |
an•an+1 |