已知公差为d的等差数列an,0<a1<π2,0<d<π2,其前n项和为Sn,若sin(a1+a3)=sina2,cos(a3-a1)=cosa2.(1)求数列an的通项公式;(2)设bn=Sn(n+1

题目简介

已知公差为d的等差数列an,0<a1<π2,0<d<π2,其前n项和为Sn,若sin(a1+a3)=sina2,cos(a3-a1)=cosa2.(1)求数列an的通项公式;(2)设bn=Sn(n+1

题目详情

已知公差为d的等差数列an,0<a1
π
2
,0<d<
π
2
,其前n项和为Sn,若sin(a1+a3)=sina2,cos(a3-a1)=cosa2
(1)求数列an的通项公式;
(2)设bn=
Sn
(n+1)•2n-1
,求数列bn的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵sin(a1+a3)=sina2,∴sin2a2=2sina2cosa2=sina2,∴sina2(2cosa2-1)=0,
∵0<a1<class="stub"π
2
,0<d<class="stub"π
2
,∴0<a2<π,∴sina2≠0,∴cosa2=class="stub"1
2
,∴a2=class="stub"π
3

∵cos(a3-a1)=cosa2,∴cos2d=cosclass="stub"π
3
,∴d=class="stub"π
6
,∴a1=class="stub"π
6
,∴an=class="stub"π
6
+
(n-1)•class="stub"π
6
=class="stub"nπ
6
,∴数列an的通项公式为an=class="stub"nπ
6

(2)∵Sn=
n(a1+an)
2
=
n(n+1)π
12
,∴bn=
Sn
(n+1)•2n-1
=class="stub"πn
6•2n
=class="stub"π
6
•class="stub"n
2n

Tn=class="stub"π
6
(class="stub"1
2
+2•class="stub"1
22
+3•class="stub"1
23
+4•class="stub"1
24
++n•class="stub"1
2n
)
①,class="stub"1
2
Tn=class="stub"π
6
[class="stub"1
22
+2•class="stub"1
23
+3•class="stub"1
24
+4•class="stub"1
25
++(n-1)•class="stub"1
2n
+n•class="stub"1
2n+1
]
②,
①-②得class="stub"1
2
Tn=class="stub"π
6
(class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+class="stub"1
24
++class="stub"1
2n
-n•class="stub"1
2n+1
)
=
class="stub"π
12
(1-class="stub"1
2n
)
1-class="stub"1
2
-class="stub"nπ
6
•class="stub"1
2n+1
=class="stub"π
6
(1-class="stub"1
2n
)-class="stub"nπ
6
•class="stub"1
2n+1

Tn=class="stub"π
3
-
(n+2)π
3•2n+1

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