已知数列{an}满足a1=76,点(2Sn+an,Sn+1)在f(x)=12x+13的图象上:(1)求数列{an}的通项公式;(2)若cn=(an-23)n,Tn为cn的前n项和,求Tn.-数学

题目简介

已知数列{an}满足a1=76,点(2Sn+an,Sn+1)在f(x)=12x+13的图象上:(1)求数列{an}的通项公式;(2)若cn=(an-23)n,Tn为cn的前n项和,求Tn.-数学

题目详情

已知数列{an}满足a1=
7
6
,点(2Sn+an,Sn+1)在f(x)=
1
2
x+
1
3
的图象上:
(1)求数列{an}的通项公式;
(2)若cn=(an-
2
3
)n,Tn为cn的前n项和,求Tn
题型:解答题难度:中档来源:不详

答案

(1)解∵点(2Sn+an,Sn+1)在f(x)=class="stub"1
2
x+class="stub"1
3
的图象上
Sn+1=class="stub"1
2
(2Sn+an)+class="stub"1
3
Sn+1-Sn=class="stub"1
2
an+class="stub"1
3

an+1=class="stub"1
2
an+class="stub"1
3

an+1-class="stub"2
3
=class="stub"1
2
(an-class="stub"2
3
)

∴{an-class="stub"2
3
}是以a1-class="stub"2
3
=class="stub"1
2
为首项,以class="stub"1
2
为公比的等比数列
an-class="stub"2
3
=class="stub"1
2
•(class="stub"1
2
)n-1

an=class="stub"2
3
+(class="stub"1
2
)n

(2)∵cn=(an-class="stub"2
3
)n=class="stub"n
2n

Tn=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n
2n
 …①
class="stub"1
2
Tn=class="stub"1
22
+class="stub"2
23
+…+class="stub"n
2n+1
.②
①-②得class="stub"1
2
Tn=class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
-class="stub"n
2n+1

Tn=2-class="stub"1
2n-1
-class="stub"n
2n

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