(2009•崇文区一模)已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn}满足条件:a1=1,a
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(2009•崇文区一模)已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn}满足条件:a1=1,an+1=g(an)+1(n∈N*),b
(Ⅰ)由题意an+1=2an+1,
∴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.
∴an+1=2×2n-1,
∴an=2n-1.
(Ⅱ)∵bn=
1
(2n+1)(2n+3)=
1
2(
1
2n+1−
1
2n+3),
∴Tn=
1
2(
1
3−
1
5+
1
5−
1
7++
1
2n+1−
1
2n+3)=
1
2(
1
3−
1
2n+3)=
n
3×(2n+3)=
n
6n+9.
∵
Tn+1
Tn=
n+1
6n+15•
6n+9
n=
6n2+15n+9
6n2+15n>1,
∴Tn<Tn+1,n∈N*.
∴当n=1时,Tn取得最小值
1
15.
由题意得
1
15>
m
150,
∴m<10.
∵m∈Z,
∴m=9.
(Ⅲ)证明:∵
ak
ak+1=
2k−1
2k+1−1=
2k−1
2(2k−
1
2)<
1
2,k=1,2,3,,n,
∴
a1
a2+
a2
a3++
∴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.
∴an+1=2×2n-1,
∴an=2n-1.
(Ⅱ)∵bn=
1
(2n+1)(2n+3)=
1
2(
1
2n+1−
1
2n+3),
∴Tn=
1
2(
1
3−
1
5+
1
5−
1
7++
1
2n+1−
1
2n+3)=
1
2(
1
3−
1
2n+3)=
n
3×(2n+3)=
n
6n+9.
∵
Tn+1
Tn=
n+1
6n+15•
6n+9
n=
6n2+15n+9
6n2+15n>1,
∴Tn<Tn+1,n∈N*.
∴当n=1时,Tn取得最小值
1
15.
由题意得
1
15>
m
150,
∴m<10.
∵m∈Z,
∴m=9.
(Ⅲ)证明:∵
ak
ak+1=
2k−1
2k+1−1=
2k−1
2(2k−
1
2)<
1
2,k=1,2,3,,n,
∴
a1
a2+
a2
a3++
(2009•崇文区一模)已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn}满足条件:a1=1,a
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