已知等差数列{an}的首项a1=3,且公差d≠0,其前n项和为Sn,且a1,a4,a13分别是等比数列{bn}的b2,b3,b4.(Ⅰ)求数列{an}与{bn}的通项公式;(Ⅱ)证明13≤1S1+1S

题目简介

已知等差数列{an}的首项a1=3,且公差d≠0,其前n项和为Sn,且a1,a4,a13分别是等比数列{bn}的b2,b3,b4.(Ⅰ)求数列{an}与{bn}的通项公式;(Ⅱ)证明13≤1S1+1S

题目详情

已知等差数列{an}的首项a1=3,且公差d≠0,其前n项和为Sn,且a1,a4,a13分别是等比数列{bn}的b2,b3,b4
(Ⅰ)求数列{an}与{bn}的通项公式;
(Ⅱ)证明
1
3
1
S1
+
1
S2
+…+
1
Sn
3
4
题型:解答题难度:中档来源:泰安二模

答案

(Ⅰ)设等比数列的公比为q,则
∵a1,a4,a13分别是等比数列{bn}的b2,b3,b4.
(a1+3d)2=a1(a1+12d)
∵a1=3,∴d2-2d=0
∴d=2或d=0(舍去)
∴an=3+2(n-1)=2n+1
q=
b3
b2
=
a4
a1
=3
b1=
b2
q
=1

∴bn=3n-1;
(Ⅱ)证明:由(Ⅰ)知Sn=n2+2n
class="stub"1
Sn
=class="stub"1
n(n+2)
=class="stub"1
2
class="stub"1
n
-class="stub"1
n+2

class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
Sn
=class="stub"1
2
[(1-class="stub"1
3
)+(class="stub"1
2
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+2
)]
=class="stub"1
2
(1+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2
)

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

class="stub"1
n+1
+class="stub"1
n+2
class="stub"1
2
+class="stub"1
3
=class="stub"5
6

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

class="stub"1
3
≤class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
Sn
<class="stub"3
4

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