题目
题型:安徽模拟难度:来源:
2an |
an+2 |
1 |
1006 |
(Ⅰ)求证:数列{
1 |
an |
(Ⅱ)若bn=
2-2010an |
an |
1 |
2 |
(Ⅲ)比较Tn与
5n |
2n+1 |
答案
2an |
an+2 |
1 |
an+1 |
1 |
an |
1 |
2 |
数列{
1 |
an |
1 |
a1 |
1 |
2 |
故
1 |
an |
1 |
a1 |
1 |
2 |
2+(n-1)a1 |
2a1 |
因为a1=
1 |
1006 |
所以数列{xn}的通项公式为an=
2a1 |
(n-1)a1+2 |
2 |
n+2011 |
(Ⅱ)将an代入bn可求得bn=
2-2010×
| ||
|
所以cn=bn•(
1 |
2 |
1 |
2 |
Tn=2×
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
由①-②得
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
=1+
| ||||
1-
|
1 |
2 |
3 |
2 |
n+3 |
2n+1 |
∴Tn=3-
n+3 |
2n |
(Ⅲ)Tn-
5n |
2n+1 |
n+3 |
2n |
5n |
2n+1 |
(n+3)(2n-2n-1) |
2n(2n+1) |
于是确定Tn与
5n |
2n+1 |
当n=1时,Tn=3-
n+3 |
2n |
5n |
2n+1 |
5 |
3 |
5n |
2n+1 |
当n=2时,Tn=3-
n+3 |
2n |
5 |
4 |
7 |
4 |
5n |
2n+1 |
5n |
2n+1 |
当n=3时,23=8>2×3+1=7,
当n=4时,24=16>2×4+1=9,
…
可猜想当n≥3时,2n>2n+1…(11分)
证明如下:
(1)当n=3时,由上验算显示成立,
(2)假设n=k时成立,即2k>2k+1
则n=k+1时2•2k>2(2k+1)=4k+2=2(k+1)+1+(2k-1)>2(k+1)+1
所以当n=k+1时猜想也成立
综合(1)(2)可知,对一切n≥3的正整数,都有2n>2n+1…(12分)
综上所述,当n=1,2时,Tn<
5n |
2n+1 |
当n≥3时,Tn>
5n |
2n+1 |
核心考点
试题【已知数列{an}满足2anan+2an+1(n∈N*),且a1=11006.(Ⅰ)求证:数列{1an}是等差数列,并求通项an;(Ⅱ)若bn=2-2010ana】;主要考察你对等差数列等知识点的理解。[详细]
举一反三
1 |
2 |
1 |
2 |
(Ⅰ)令bn=2nan,求证数列{bn}是等差数列,并求数列{an}的通项公式;
(Ⅱ)令cn=
n+1 |
n |
A.27 | B.6 | C.81 | D.9 |