数列{an}的前n项和为Sn,已知Sn+an=-n(n∈N*)恒成立.(1)求数列{an}的通项公式;(2)bn=ln(an+1),求{anbn}的前n项和;(3)求证:12a1a2+122a2a3+

题目简介

数列{an}的前n项和为Sn,已知Sn+an=-n(n∈N*)恒成立.(1)求数列{an}的通项公式;(2)bn=ln(an+1),求{anbn}的前n项和;(3)求证:12a1a2+122a2a3+

题目详情

数列{an}的前n项和为Sn,已知Sn+an=-n(n∈N*)恒成立.
(1)求数列{an}的通项公式;
(2)bn=ln(an+1),求{anbn}的前n项和;
(3)求证:
1
2a1a2
+
1
22a2a3
+…+
1
2nanan+1
<2
题型:解答题难度:中档来源:香洲区模拟

答案

(1)∵Sn+an=-n①
∴n≥2时,Sn-1+an-1=-n+1②
①-②可得2an=an-1-1
∴2(an+1)=an-1+1
又a1=-class="stub"1
2
,∴{an+1}是以class="stub"1
2
为首项,class="stub"1
2
为公比的等比数列
∴an+1=(class="stub"1
2
)n
,∴an=(class="stub"1
2
)
n
-1;
(2)bn=ln(an+1)=nlnclass="stub"1
2
,∴anbn=[(class="stub"1
2
)
n
-1]•nlnclass="stub"1
2

∴{anbn}的前n项和为lnclass="stub"1
2
[class="stub"1
2
+2•(class="stub"1
2
)
2
+…+n•(class="stub"1
2
)
n
]-
n(n+1)
2
•lnclass="stub"1
2

令Tn=lnclass="stub"1
2
[class="stub"1
2
+2•(class="stub"1
2
)
2
+…+n•(class="stub"1
2
)
n
],则class="stub"1
2
Tn=lnclass="stub"1
2
[(class="stub"1
2
)
2
+2•(class="stub"1
2
)
3
+…+(n-1)•(class="stub"1
2
)
n
+n•(class="stub"1
2
)
n+1
],
两式相减,可得Tn=lnclass="stub"1
2
(2-class="stub"1
2n-1
-class="stub"n
2n

∴{anbn}的前n项和为lnclass="stub"1
2
(2-class="stub"1
2n-1
-class="stub"n
2n
)-
n(n+1)
2
•lnclass="stub"1
2

(3)证明:由(1)知,class="stub"1
2nanan+1
=-2(class="stub"1
class="stub"1
2n
-1
-class="stub"1
class="stub"1
2n+1
-1

class="stub"1
2a1a2
+class="stub"1
22a2a3
+…+class="stub"1
2nanan+1
=-2(class="stub"1
class="stub"1
21
-1
-class="stub"1
class="stub"1
22
-1
+class="stub"1
class="stub"1
22
-1
-class="stub"1
class="stub"1
3
-1
+…+class="stub"1
class="stub"1
2n
-1
-class="stub"1
class="stub"1
2n+1
-1

=-2(class="stub"1
class="stub"1
21
-1
-class="stub"1
class="stub"1
2n+1
-1
)<2
class="stub"1
2a1a2
+class="stub"1
22a2a3
+…+class="stub"1
2nanan+1
<2

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