设数列{an}的前n项和为Sn,已知数列{an}的各项都为正数,且对任意n∈N*都有2pSn=an2+pan(其中p>0为常数)(1)求数列{an}的通项公式;(2)若对任意n∈N*都有1S1+1S2

题目简介

设数列{an}的前n项和为Sn,已知数列{an}的各项都为正数,且对任意n∈N*都有2pSn=an2+pan(其中p>0为常数)(1)求数列{an}的通项公式;(2)若对任意n∈N*都有1S1+1S2

题目详情

设数列{an}的前n项和为Sn,已知数列{an}的各项都为正数,且对任意n∈N*都有2pSn=an2+pan(其中p>0为常数)
(1)求数列{an}的通项公式;
(2)若对任意n∈N*都有
1
S1
+
1
S2
+
…+
1
Sn
<1成立,求p的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,2pS1=a12+pa1,∴a1=p,
∵2pSn=an2+pan,∴n≥2时,2pSn-1=an-12+pan-1,
两式相减可得p(an+an-1)=(an-an-1)(an+an-1)
∵an+an-1>0,∴an-an-1=p
∴数列{an}是首项和公差都为p的等差数列
∴an=np;
(2)由(1)知Sn=
n(p+np)
2
,∴class="stub"1
Sn
=class="stub"2
p
(class="stub"1
n
-class="stub"1
n+1
)

class="stub"1
S1
+
class="stub"1
S2
+
…+class="stub"1
Sn
=class="stub"2
p
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)=class="stub"2
p
(1-class="stub"1
n+1
)
=class="stub"2
p
•class="stub"n
n+1

∵对任意n∈N*都有class="stub"1
S1
+
class="stub"1
S2
+
…+class="stub"1
Sn
<1成立,
class="stub"2
p
•class="stub"n
n+1
<1

class="stub"n
n+1
<class="stub"p
2

class="stub"n
n+1
<1

class="stub"p
2
≥1,即p≥2.

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