题目
题型:不详难度:来源:
(1)求证:数列{an-2n}为等差数列;
(2)设数列{bn}满足bn=log2(an+1-n),若(1+
1 |
b2 |
1 |
b3 |
1 |
b4 |
1 |
bn |
n+1 |
答案
故数列{an-2n}为等差数列,且公差d=1.
an-2n=(a1-2)+(n-1)d=n-1,an=2n+n-1;
(2)由(1)可知an=2n+n-1,∴bn=log2(an+1-n)=n
设f(n)=(1+
1 |
b2 |
1 |
b3 |
1 |
b4 |
1 |
bn |
1 | ||
|
则f(n+1)=(1+
1 |
b2 |
1 |
b3 |
1 |
b4 |
1 |
bn |
1 |
bn+1 |
1 | ||
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两式相除可得
f(n+1) |
f(n) |
1 |
bn+1 |
| ||
|
| ||
|
则有f(n)>f(n-1)>f(n-2)>…>f(2)=
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2 |
要使(1+
1 |
b2 |
1 |
b3 |
1 |
b4 |
1 |
bn |
n+1 |
必有k<
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2 |
故k的取值范围是k<
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2 |
核心考点
试题【在数列{an}中,a1=2,an+1=an+2n+1(n∈N*)(1)求证:数列{an-2n}为等差数列;(2)设数列{bn}满足bn=log2(an+1-n)】;主要考察你对等差数列等知识点的理解。[详细]
举一反三
a | 24 |