题目内容
对于数列{an},规定{△an}为数列{an}的一阶差分数列,其中△an=an+1-an(n∈N*);一般地,规定{△kan}为数列{an}的k阶差分数列,其中△kan=△k-1an+1-△k-1an,且k∈N*,k≥2.(Ⅰ)已知数列{an}的通项公式an=
5 |
2 |
13 |
2 |
(Ⅱ)若数列{an}的首项a1=1,且满足△2an-an+1+an=-2n(n∈N*),求数列{an}的通项公式;
(Ⅲ)在(Ⅱ)的条件下,记bn=
|
b2 |
2 |
bn |
n |
17 |
12 |
分析:(Ⅰ)根据题意:△an=an+1-an=(n+1)2-(n+1)-n2+n=5n-4,所以△an+1-△an=6.由此能够证明{△an}是等差数列.
(Ⅱ)由△2an-△an+1+an=-2n,知△an+1-△an-△an+1+an=-2n,所以△an-an=2n.由此入手能够求出数列{an}的通项公式.
(Ⅲ)由an=n•2n-1,bn=
=
=
,当n≥2,n∈N*时,
=
=
(
-
),由此入手,能够证明b1+
+…+
<
.
(Ⅱ)由△2an-△an+1+an=-2n,知△an+1-△an-△an+1+an=-2n,所以△an-an=2n.由此入手能够求出数列{an}的通项公式.
(Ⅲ)由an=n•2n-1,bn=
|
|
|
bn |
n |
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
b2 |
2 |
bn |
n |
17 |
12 |
解答:解:(Ⅰ)根据题意:△an=an+1-an=(n+1)2-(n+1)-n2+n=5n-4 (2分)
∴△an+1-△an=6.
∴数列{Dan}是首项为1,公差为5的等差数列.(3分)
(Ⅱ)由△2an-△an+1+an=-2n,∴△an+1-△an-△an+1+an=-2n,?△an-an=2n.(5分)
而△an=an+1-an,∴an+1-2an=2n,∴
-
=
,(6分)
∴数列{
}构成以
为首项,
为公差的等差数列,
即
=
?an=n•2n-1.(7分)
(Ⅲ)由(Ⅱ)知an=n•2n-1,
∴bn=
=
=
(9分)
∴当n≥2,n∈N*时
=
=
(
-
),
∴b1+
+…+
=1+[(
-
)+(
-
)+(
-
)+…+(
-
)+(
-
)]
=1+
(
+
-
-
)<1+
(
+
)=
.
当n=1时,b1=1<
,显然成立.
∴b1+
+…+
<
.(12分)
∴△an+1-△an=6.
∴数列{Dan}是首项为1,公差为5的等差数列.(3分)
(Ⅱ)由△2an-△an+1+an=-2n,∴△an+1-△an-△an+1+an=-2n,?△an-an=2n.(5分)
而△an=an+1-an,∴an+1-2an=2n,∴
an+1 |
2n+1 |
an |
2n |
1 |
2 |
∴数列{
an |
2n |
1 |
2 |
1 |
2 |
即
an |
2n |
n |
2 |
(Ⅲ)由(Ⅱ)知an=n•2n-1,
∴bn=
|
|
|
∴当n≥2,n∈N*时
bn |
n |
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
∴b1+
b2 |
2 |
bn |
n |
1 |
2 |
1 |
4 |
1 |
3 |
1 |
5 |
1 |
4 |
1 |
6 |
1 |
n-1 |
1 |
n+1 |
1 |
n |
1 |
n+2 |
=1+
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n+1 |
1 |
n+2 |
1 |
2 |
1 |
2 |
1 |
3 |
17 |
12 |
当n=1时,b1=1<
17 |
12 |
∴b1+
b2 |
2 |
bn |
n |
17 |
12 |
点评:第(Ⅰ)题考查等差数列的证明,解题时要注意等差数列性质的合理运用;第(Ⅱ)题考查数列通项公式的求解方法,解题时要注意构造法的合理运用;第(Ⅲ)题考查数列前n项和的证明,解题时要注意裂项求和法的合理运用.
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