等差数列{an}中,a1=3,前n项和为Sn,等比数列{bn}各项均为正数,b1=1,且b2+S2=12,{bn}的公比q=S2S1.(1)求an与bn.(2)证明:13≤1S1+1S2+…+1Sn小

题目简介

等差数列{an}中,a1=3,前n项和为Sn,等比数列{bn}各项均为正数,b1=1,且b2+S2=12,{bn}的公比q=S2S1.(1)求an与bn.(2)证明:13≤1S1+1S2+…+1Sn小

题目详情

等差数列{an}中,a1=3,前n项和为Sn,等比数列{bn}各项均为正数,b1=1,且b2+S2=12,{bn}的公比q=
S2
S1

(1)求an与bn
(2)证明:
1
3
1
S1
+
1
S2
+…+
1
Sn
小于
2
3
题型:解答题难度:中档来源:不详

答案

(I)由已知可得
q+3+a2=12
q=
3+a2
q

解得,q=3或q=-4(舍去),a2=6
∴an=3+(n-1)3=3n
∴bn=3n-1
(2)证明:∵Sn=
n(3+3n)
2
class="stub"1
Sn
=class="stub"2
n(3+3n)
=class="stub"2
3
(class="stub"1
n
-class="stub"1
n+1
)

class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
Sn

=class="stub"2
3
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+…+class="stub"1
n
-class="stub"1
n+1
)

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

∵n≥1∴0<class="stub"1
n+1
≤class="stub"1
2
class="stub"1
3
≤class="stub"2
3
(1-class="stub"1
n+1
)<class="stub"2
3

class="stub"1
3
≤class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
Sn
<class="stub"2
3

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