数列{an}前n项和为Sn=n2+2n,等比数列{bn}各项为正数,且b1=1,{ban}是公比为64的等比数列.(1)求数列{an}与{bn}的通项公式;(2)证明:1S1+1S2+…+1Sn<34

题目简介

数列{an}前n项和为Sn=n2+2n,等比数列{bn}各项为正数,且b1=1,{ban}是公比为64的等比数列.(1)求数列{an}与{bn}的通项公式;(2)证明:1S1+1S2+…+1Sn<34

题目详情

数列{an}前n项和为Sn=n2+2n,等比数列{bn}各项为正数,且b1=1,{ban}是公比为64的等比数列.
(1)求数列{an}与{bn}的通项公式;
(2)证明:
1
S1
+
1
S2
+…+
1
Sn
3
4
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,a1=S1=3,
n≥2时,an=Sn-Sn-1=(n2+2n)-{(n-1)2+2(n-1)}=2n+1
经验证,当n=1时,上式也适合,故an=2n+1.
设{bn}公比为q,则
ba2
ba1
=
b5
b3
=q2=64

因为{bn}各项为正数所以q=8,∴bn=8n-1
故数列{an}与{bn}的通项公式分别为:an=2n+1,bn=8n-1
(2)由题意可知class="stub"1
Sn
=class="stub"1
n2+2n
=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
3
-class="stub"1
5
+…+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

故原不等式得证.

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