已知数列{an}的前n项和为Sn,其中a1=12,5Sn=7an-an-1+5Sn-1(n≥2);等差数列{bn},其中b3=2,b5=6,.(1)求数列{an}的通项公式;(2)若cn=(bn+3)

题目简介

已知数列{an}的前n项和为Sn,其中a1=12,5Sn=7an-an-1+5Sn-1(n≥2);等差数列{bn},其中b3=2,b5=6,.(1)求数列{an}的通项公式;(2)若cn=(bn+3)

题目详情

已知数列{an}的前n项和为Sn,其中a1=
1
2
,5Sn=7an-an-1+5Sn-1(n≥2);等差数列{bn},其中b3=2,b5=6,.
(1)求数列{an}的通项公式;
(2)若cn=(bn+3)an,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵5Sn=7an-an-1+5Sn-1(n≥2);
∴5Sn-5Sn-1=7an-an-1,
∴2an=an-1,
an
an-1
=class="stub"1
2

即数列{an}是公比q=class="stub"1
2
的等比数列,
a1=class="stub"1
2
,∴an=class="stub"1
2
(class="stub"1
2
)n-1=class="stub"1
2n

(2)在等差数列{bn},
∵b3=2,b5=6,
b1+2d=2
b1+4d=6
,解得
b1=-2
d=2

∴bn=-2+2(n-1)=2n-4,
∵cn=(bn+3)an,
∴cn=(bn+3)an=(2n-1)•class="stub"1
2n

Tn=1×class="stub"1
2
+3×class="stub"1
22
+5×class="stub"1
23
+…+(2n-1)×class="stub"1
2n
class="stub"1
2
Tn=1×class="stub"1
22
+3×class="stub"1
23
+…+(2n-3)×class="stub"1
2n
+(2n-1)×class="stub"1
2n+1

两式作差得:
class="stub"1
2
Tn=class="stub"1
2
+2×(class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
)-(2n-1)×class="stub"1
2n+1
=class="stub"1
2
+
2×class="stub"1
22
[1-(class="stub"1
2
)
n-1
]
1-class="stub"1
2
-(2n-1)×class="stub"1
2n+1

Tn=3-class="stub"2n+3
2n

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