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=S2b2(1)求an与bn;(2)求1S1+1S2+…+1Sn.-数学

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=S2b2(1)求an与bn;(2)求1S1+1S2+…+1Sn.-数学

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
b2

(1)求an与bn
(2)求
1
S1
+
1
S2
+…+
1
Sn
题型:解答题难度:中档来源:花都区模拟

答案

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

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