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等差数列{an}的前n项和为Sn,数列{bn}是等比数列,满足a1=2,b1=1,b2+S2=8,a5-2b2=a3.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)令cn=an,n为奇数bn,n为偶数,设数列{cn}前n项和为Tn,求T2n.
题目详情
等差数列{an}的前n项和为Sn,数列{bn}是等比数列,满足a1=2,b1=1,b2+S2=8,a5-2b2=a3.
(Ⅰ)求数列{an}和{bn}的通项公式;
(Ⅱ)令cn=
,设数列{cn}前n项和为Tn,求T2n.nnn1122523
nn
cn=
,设数列{cn}前n项和为Tn,求T2n.cn=
,设数列{cn}前n项和为Tn,求T2n.cn=
,设数列{cn}前n项和为Tn,求T2n.n=
,设数列{cn}前n项和为Tn,求T2n.
an,n为奇数 bn,n为偶数 an,n为奇数 an,n为奇数 an,n为奇数an,n为奇数n,n为奇数bn,n为偶数 bn,n为偶数 bn,n为偶数bn,n为偶数n,n为偶数 nn2n
(Ⅰ)求数列{an}和{bn}的通项公式;
(Ⅱ)令cn=
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nn
cn=
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an,n为奇数 |
bn,n为偶数 |
an,n为奇数 |
bn,n为偶数 |
an,n为奇数 |
bn,n为偶数 |
an,n为奇数 |
bn,n为偶数 |
▼优质解答
答案和解析
(Ⅰ)设数列{ann}的公差为d,数列{bnn}的公比为q,则
由
得
解得
,
∴an=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
.
b2+S2=8 a5-2b2=a3 b2+S2=8 b2+S2=8 b2+S2=82+S2=82=8a5-2b2=a3 a5-2b2=a3 a5-2b2=a35-2b2=a32=a33 得
解得
,
∴an=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
.
q+4+d=8 2+4d-2q=2+2d q+4+d=8 q+4+d=8 q+4+d=82+4d-2q=2+2d 2+4d-2q=2+2d 2+4d-2q=2+2d 解得
,
∴an=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
.
d=2 q=2 d=2 d=2 d=2q=2 q=2 q=2 ,
∴ann=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
. bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
. n=2n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
. n-1.
(Ⅱ)由(Ⅰ),cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
. cn=
,
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
+
=2n2+
•4n-
. n=
2n,n为奇数 2n-1,n为偶数 2n,n为奇数 2n,n为奇数 2n,n为奇数2n-1,n为偶数 2n-1,n为偶数 2n-1,n为偶数n-1,n为偶数 ,
∴T2n2n=(c11+c33+…+c2n-12n-1)+(c22+c44+…+c2n2n)
=[2+6+…+(4n-2)]+(2+233+…+22n-12n-1)
=
+
=2n2+
•4n-
.
n(2+4n-2) 2 n(2+4n-2) n(2+4n-2) n(2+4n-2)2 2 2+
2(1-4n) 1-4 2(1-4n) 2(1-4n) 2(1-4n)n)1-4 1-4 1-4
=2n2+
•4n-
. 2n2+
•4n-
. 2+
2 3 2 2 23 3 3•4n-
. n-
2 3 2 2 23 3 3.
由
|
|
|
∴an=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
|
b2+S2=8 |
a5-2b2=a3 |
b2+S2=8 |
a5-2b2=a3 |
b2+S2=8 |
a5-2b2=a3 |
b2+S2=8 |
a5-2b2=a3 |
|
|
∴an=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
|
q+4+d=8 |
2+4d-2q=2+2d |
q+4+d=8 |
2+4d-2q=2+2d |
q+4+d=8 |
2+4d-2q=2+2d |
q+4+d=8 |
2+4d-2q=2+2d |
|
∴an=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
|
d=2 |
q=2 |
d=2 |
q=2 |
d=2 |
q=2 |
d=2 |
q=2 |
∴ann=2+2(n-1)=2n,bn=2n-1.
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
(Ⅱ)由(Ⅰ),cn=
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
|
∴T2n=(c1+c3+…+c2n-1)+(c2+c4+…+c2n)
=[2+6+…+(4n-2)]+(2+23+…+22n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
|
2n,n为奇数 |
2n-1,n为偶数 |
2n,n为奇数 |
2n-1,n为偶数 |
2n,n为奇数 |
2n-1,n为偶数 |
2n,n为奇数 |
2n-1,n为偶数 |
∴T2n2n=(c11+c33+…+c2n-12n-1)+(c22+c44+…+c2n2n)
=[2+6+…+(4n-2)]+(2+233+…+22n-12n-1)
=
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
n(2+4n-2) |
2 |
2(1-4n) |
1-4 |
=2n2+
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
2 |
3 |
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