已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn},{cn}满足条件:a1=1,an=f(bn)=g(bn+1)(n∈N*),cn=1[12f(n)+12][g(n)+3].(

题目简介

已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn},{cn}满足条件:a1=1,an=f(bn)=g(bn+1)(n∈N*),cn=1[12f(n)+12][g(n)+3].(

题目详情

已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn},{cn}满足条件:a1=1,an=f(bn)=g(bn+1)(n∈N*),cn=
1
[
1
2
f(n)+
1
2
][g(n)+3]

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)求数列{cn}的前n项和Tn,并求使得Tn
m
150
对任意n∈N*都成立的最大正整数m;
(Ⅲ)求证:
a1
a2
+
a2
a3
+…+
an
an+1
n
2
-
1
3
题型:解答题难度:中档来源:崇文区一模

答案

(Ⅰ)由题意an+1=4bn+1+1,an=2bn+1,
∴an+1=2an+1,(2分)
∴an+1+1=2(an+1),
∵a1=1,
∴数列{an+1}是首项为2,公比为2的等比数列.(4分)
∴.an+1=2×2n-1
∴an=2n-1.(5分)
(Ⅱ)∵cn=class="stub"1
(2n+1)(2n+3)
=class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+3
)
,(7分)
Tn=class="stub"1
2
(class="stub"1
3
-class="stub"1
5
+class="stub"1
5
-class="stub"1
7
++class="stub"1
2n+1
-class="stub"1
2n+3
)
=class="stub"1
2
(class="stub"1
3
-class="stub"1
2n+3
)=class="stub"n
3×(2n+3)
=class="stub"n
6n+9
.(8分)
Tn+1
Tn
=class="stub"n+1
6n+15
•class="stub"6n+9
n
=
6n2+15n+9
6n2+15n
>1

∴Tn<Tn+1,n∈N*.
∴当n=1时,Tn取得最小值class="stub"1
15
.(10分)
由题意得class="stub"1
15
>class="stub"m
150
,得m<10.
∵m∈Z,
∴由题意得m=9.(11分)
(Ⅲ)证明:
ak
ak+1
=
2k-1
2k+1-1
=class="stub"1
2
-class="stub"1
2(2k+1-1)
=class="stub"1
2
-class="stub"1
2k+2k-2
≥class="stub"1
2
-class="stub"1
3
•class="stub"1
2k

k=1,2,3,,n(12分)
a1
a2
+
a2
a3
++
an
an+1
≥class="stub"n
2
-class="stub"1
3
(class="stub"1
2
+class="stub"1
22
++class="stub"1
2n
)=class="stub"n
2
-class="stub"1
3
(1-class="stub"1
2n
)

a1
a2
+
a2
a3
++
an
an+1
>class="stub"n
2
-class="stub"1
3
(n∈N*).(14分)

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