题目
题型:不详难度:来源:
(1)求数列{an}的通项公式;
(2)假设bn=
an |
(an+1)(an+1+1) |
127 |
390 |
答案
∴2(a3+2)=a2+a4,a3=8,a2+a4=80,
∴
|
解得a1=2,q=2,或a1=32,q=
1 |
2 |
∴an=2n.
(2)bn=
an |
(an+1)(an+1+1) |
=
2n |
(2n+1)(2n+1+1) |
=
1 |
2n+1 |
1 |
2n+1+1 |
∴Tn=
1 |
2+1 |
1 |
22+1 |
1 |
22+1 |
1 |
23-1 |
1 |
2n-1+1 |
1 |
2n+1 |
1 |
2n+1 |
1 |
2n+1+1 |
=
1 |
2+1 |
1 |
2n+1+1 |
=
1 |
3 |
1 |
2n+1+1 |
∵Tn<
127 |
390 |
∴
1 |
3 |
1 |
2n+1+1 |
127 |
130 |
∴不等式Tn<
127 |
390 |
核心考点
试题【已知递增的等比数列{an}满足:a2+a3+a4=28,a3+2是a2与a4的等差中项.(1)求数列{an}的通项公式;(2)假设bn=an(an+1)(an+】;主要考察你对数列综合等知识点的理解。[详细]
举一反三
1 |
2n |
(1)求证:数列{2nan}为等差数列,并求数列{an}的通项;
(2)求数列{Sn}的前n项和Tn.
1 |
n |
nπ |
25 |
A.25 | B.50 | C.75 | D.100 |