数列{an}为等差数列,an为正整数,其前n项和为Sn,数列{bn}为等比数列,且a1=3,b1=1,数列{ban}是公比为64的等比数列,b2S2=64.(1)求an,bn;(2)求证1S1+1S2

题目简介

数列{an}为等差数列,an为正整数,其前n项和为Sn,数列{bn}为等比数列,且a1=3,b1=1,数列{ban}是公比为64的等比数列,b2S2=64.(1)求an,bn;(2)求证1S1+1S2

题目详情

数列{an}为等差数列,an为正整数,其前n项和为Sn,数列{bn}为等比数列,且a1=3,b1=1,数列{ban}是公比为64的等比数列,b2S2=64.
(1)求an,bn
(2)求证
1
S1
+
1
S2
+…+
1
Sn
3
4
题型:解答题难度:中档来源:江西

答案

(1)设{an}的公差为d,{bn}的公比为q,则d为正整数,an=3+(n-1)d,bn=qn-1
依题意有
ban+1
ban
=
q3+nd
q3+(n-1)d
=qd=64=26
S2b2=(6+d)q=64

由(6+d)q=64知q为正有理数,故d为6的因子1,2,3,6之一,
解①得d=2,q=8
故an=3+2(n-1)=2n+1,bn=8n-1
(2)Sn=3+5++(2n+1)=n(n+2)
class="stub"1
S1
+class="stub"1
S2
++class="stub"1
Sn
=class="stub"1
1×3
+class="stub"1
2×4
+class="stub"1
3×5
++class="stub"1
n(n+2)
=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

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