题目内容
(2011•江西模拟)已知数列{an}满足an+1=
(n∈N*),a2011=
.
(1)求{an}的通项公式;
(2)若bn=
-4023且cn=
(n∈N*),求证:c1+c2+…+cn<n+1.
2an |
an+2 |
1 |
2011 |
(1)求{an}的通项公式;
(2)若bn=
4 |
an |
| ||||
2bn+1bn |
分析:(1)由已知,得
-
=
(n∈N*),从而数列{
}是以
为首项,
为公差的等差数列,然后表示出{an}的通项公式,根据a2011=
可求出a1,从而求出{an}的通项公式;
(2)将an=
代入可得bn=4×
-4023=2n-1然后求出cn,然后计算c1+c2+…+cn-n,经过化简可证得结论.
1 |
an+1 |
1 |
an |
1 |
2 |
1 |
an |
1 |
a1 |
1 |
2 |
1 |
2011 |
(2)将an=
2 |
n+2011 |
n+2011 |
2 |
解答:解:(1)由已知,得
=
+
,即
-
=
(n∈N*),
∴数列{
}是以
为首项,
为公差的等差数列.
=
+(n-1)×
=
,
∴an=
…(4分)
又因为a2011=
=
解得a1=
∴an=
=
…(6分)
(2)证明:∵an=
,
∴bn=4×
-4023=2n-1-------(7分)
∴cn=
=
=
=1+
=1+
-
∴c1+c2+…cn-n=(1+1-
)+(1+
-
)+…+(1+
-
)-n=1-
<1
故c1+c2+…+cn<n+1…(12分)
1 |
an+1 |
1 |
2 |
1 |
an |
1 |
an+1 |
1 |
an |
1 |
2 |
∴数列{
1 |
an |
1 |
a1 |
1 |
2 |
1 |
an |
1 |
a1 |
1 |
2 |
(n-1)a1+2 |
2a1 |
∴an=
2a1 |
(n-1)a1+2 |
又因为a2011=
2a1 |
2010a1+2 |
1 |
2011 |
解得a1=
1 |
1006 |
∴an=
2×
| ||
(n-1)×
|
2 |
n+2011 |
(2)证明:∵an=
2 |
n+2011 |
∴bn=4×
n+2011 |
2 |
∴cn=
| ||||
2bn+1bn |
(2n+1)2+(2n-1)2 |
2(2n+1)(2n-1) |
4n2+1 |
4n2-1 |
2 |
(2n-1)(2n+1) |
1 |
2n-1 |
1 |
2n+1 |
∴c1+c2+…cn-n=(1+1-
1 |
3 |
1 |
3 |
1 |
5 |
1 |
2n-1 |
1 |
2n+1 |
1 |
2n+1 |
故c1+c2+…+cn<n+1…(12分)
点评:本题主要考查了构造新数列,以及等差数列的通项公式和数列的裂项求和法,同时考查了计算能力,属于中档题.
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