已知数列{an}的前n项和Sn=n(n+2),数列{bn}的前n项和为Tn,且有Tn+1-bn+1Tn+bn=1,b1=3.(1)求数列{an},{bn}的通项an,bn;(2)设cn=anbn,试判

题目简介

已知数列{an}的前n项和Sn=n(n+2),数列{bn}的前n项和为Tn,且有Tn+1-bn+1Tn+bn=1,b1=3.(1)求数列{an},{bn}的通项an,bn;(2)设cn=anbn,试判

题目详情

已知数列{an}的前n项和Sn=n(n+2),数列{bn}的前n项和为Tn,且有
Tn+1-bn+1
Tn+bn
=1,b1=3

(1)求数列{an},{bn}的通项an,bn
(2)设cn=
an
bn
,试判断数列{cn}的单调性,并证明你的结论.
(3)在(2)的前提下,设Mn是数列{cn}的前n项和,证明:Mn≥4-
n+2
2n-1
题型:解答题难度:中档来源:乐山模拟

答案

(1)∵Sn=n(n+2),
∴当n≥2时,an=Sn-Sn-1=2n+1
当n=1时,a1=S1=3满足上式
∴an=2n+1
Tn+1-bn+1
Tn+bn
=1

∴Tn+1-Tn=2bn-1
∴bn+1=2bn-1
∴bn+1-1=2(bn-1)
∴{bn-1}是公比为2的等比数列
bn-1=(b1-1)•2n-1=2n
bn =2n+1
(2)cn=
an
bn
=class="stub"2n+1
2n+1
,数列{cn}为递减数列
证明:∵cn+1-cn=class="stub"2n+3
2n+1+1
-class="stub"2n+1
2n+1

=
(1-2n)•2n+2
(2n+1+1)(2n+1)
<0

∴数列{cn}为递减数列
(3)证明:∵cn=
an
bn
=class="stub"2n+1
2n+1
≥class="stub"2n
2n
=class="stub"n
2n-1

∴Mn=c1+c2+…+cn≥1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n
2n-1

rn=1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n
2n-1

class="stub"1
2
r
n
=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n
2n

①-②:class="stub"1
2
r
n
=1+class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n-1
-class="stub"n
2n
=2-class="stub"n+2
2n

rn=4-class="stub"n+2
2n-1

1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n
2n-1
=4-class="stub"n+2
2n-1

Mn≥4-class="stub"n+2
2n-1

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