(2009•崇文区一模)已知函数f(x)=4x+1,g(x)=2x,x∈R,数列{an},{bn},{cn}满足条件: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},{cn}满足条件:a1=1,an=f(bn)=g(bn+1)(n∈N*),c
(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=
1
(2n+1)(2n+3)=
1
2(
1
2n+1−
1
2n+3),(7分)
∴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.(8分)
∵
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.(10分)
由题意得
1
15>
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=
1
2−
1
2(2k+1−1)=
1
2−
1
3×2k+2k−2≥
1
2−
1
3•
1
2k
∴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=
1
(2n+1)(2n+3)=
1
2(
1
2n+1−
1
2n+3),(7分)
∴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.(8分)
∵
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.(10分)
由题意得
1
15>
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=
1
2−
1
2(2k+1−1)=
1
2−
1
3×2k+2k−2≥
1
2−
1
3•
1
2k
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