已知等比数列an=13n-1,其前n项和为Sn=nk-1ak,则Sk+1与Sk的递推关系不满足()A.Sk+1=Sk+13k+1B.Sk+1=1+13SkC.Sk+1=Sk+ak+1D.Sk+1=3S

题目简介

已知等比数列an=13n-1,其前n项和为Sn=nk-1ak,则Sk+1与Sk的递推关系不满足()A.Sk+1=Sk+13k+1B.Sk+1=1+13SkC.Sk+1=Sk+ak+1D.Sk+1=3S

题目详情

已知等比数列an=
1
3n-1
,其前n项和为Sn=
n
k-1
ak,则Sk+1与Sk的递推关系不满足(  )
A.Sk+1=Sk+
1
3k+1
B.Sk+1=1+
1
3
Sk
C.Sk+1=Sk+ak+1D.Sk+1=3Sk-3+ak+ak+1
题型:单选题难度:偏易来源:不详

答案

∵等比数列an=class="stub"1
3n-1
=31-n,
∴a1=1,a2=class="stub"1
3
,q=class="stub"1
3

∴Sn=
n
k=1
ak
=
1-class="stub"1
3n
1-class="stub"1
3
=class="stub"3
2
(1-class="stub"1
3n
),
∴Sk+1=Sk+class="stub"1
3k
,故A不成立;
Sk+1=class="stub"3
2
(1-class="stub"1
3n
)=class="stub"3
2
-class="stub"3
2
×class="stub"1
3n

=1+class="stub"1
2
-class="stub"1
2
×class="stub"1
3n-1

=1+class="stub"1
2
(1-class="stub"1
3n-1
)
=1+class="stub"1
3
Sk
,故B成立;
由数列的前n项和的定义知:Sk+1=Sk+ak+1,故C成立;
∵3Sk-3+ak+ak+1
=3×class="stub"3
2
(1-class="stub"1
3k-1
)-3+31-k+3-k

=class="stub"9
2
-class="stub"9
2
×class="stub"1
3n-1
-3+class="stub"1
3k-1
+class="stub"3
3k-1

=class="stub"3
2
-class="stub"1
2
×class="stub"1
3k-1

=class="stub"3
2
(1-class="stub"1
3k
)
=Sk+1,故D成立.
故选A.

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