题目内容
已知数列{an}满足:a1=1,an-an-1+2anan-1=0,(n∈N*,n>1)
(Ⅰ)求证数列{
}是等差数列并求{an}的通项公式;
(Ⅱ)设bn=anan+1,求证:b1+b2+…+bn<
.
(Ⅰ)求证数列{
1 |
an |
(Ⅱ)设bn=anan+1,求证:b1+b2+…+bn<
1 |
2 |
证明:(Ⅰ)an-an-1+2anan-1=0两边同除以anan-1得:
-
=2
所以数列{
}是以1为首项,2为公差的等差数列…(3分)
于是
=2n-1,an=
,(n∈N*)…(6分)
(Ⅱ)由(Ⅰ),bn=
则
b1+b2+…+bn=
+
+…+
=
(1-
+
-
+…+
-
)=
(1-
)<
…(12分)
1 |
an |
1 |
an-1 |
所以数列{
1 |
an |
于是
1 |
an |
1 |
2n-1 |
(Ⅱ)由(Ⅰ),bn=
1 |
(2n-1)(2n+1) |
b1+b2+…+bn=
1 |
1×3 |
1 |
3×5 |
1 |
(2n-1)(2n+1) |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
2n-1 |
1 |
2n+1 |
1 |
2 |
1 |
2n+1 |
1 |
2 |
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