已知正项数列{an}的前n项和为Sn,且a1=2,4Sn=an?an+1,n∈N*.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设
已知正项数列{an}的前n项和为Sn,且a1=2,4Sn=an?an+1,n∈N*.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设数列{1an2}与的前n项和为Tn,求证:n4...
已知正项数列{an}的前n项和为Sn,且a1=2,4Sn=an?an+1,n∈N*.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设数列{1an2}与的前n项和为Tn,求证:n4n+4<Tn<12.
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(Ⅰ)解:∵4Sn=an?an+1,n∈N* ①,
∴4a1=a1?a2,
又a1=2,
∴a2=4.
当n≥2时,4Sn-1=an-1?an ②,
①-②得:4an=an?an+1-an-1?an,
由题意知an≠0,
∴an+1-an-1=4,
当n=2k+1,k∈N*时,a2k+2-a2k=4,
即a2,a4,…,a2k是首项为4,公差为4的等差数列,
∴a2k=4+4(k-1)=4k=2×2k;
当n=2k,k∈N*时,a2k+1-a2k-1=4,
即a1,a3,…,a2k-1是首项为2,公差为4的等差数列,
∴a2k-1=2+4(k-1)=4k-2=2×(2k-1).
综上可知,an=2n,n∈N*;
(Ⅱ)证明:∵
=
>
=
(
?
),
∴Tn=
+
+…+
>
(1?
+
?
+…+
?
)
=
(1?
)=
.
又∵
=
<
=
=
(
?
)
∴Tn=
∴4a1=a1?a2,
又a1=2,
∴a2=4.
当n≥2时,4Sn-1=an-1?an ②,
①-②得:4an=an?an+1-an-1?an,
由题意知an≠0,
∴an+1-an-1=4,
当n=2k+1,k∈N*时,a2k+2-a2k=4,
即a2,a4,…,a2k是首项为4,公差为4的等差数列,
∴a2k=4+4(k-1)=4k=2×2k;
当n=2k,k∈N*时,a2k+1-a2k-1=4,
即a1,a3,…,a2k-1是首项为2,公差为4的等差数列,
∴a2k-1=2+4(k-1)=4k-2=2×(2k-1).
综上可知,an=2n,n∈N*;
(Ⅱ)证明:∵
1 |
an2 |
1 |
4n2 |
1 |
4n(n+1) |
1 |
4 |
1 |
n |
1 |
n+1 |
∴Tn=
1 |
a12 |
1 |
a22 |
1 |
an2 |
1 |
4 |
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n |
1 |
n+1 |
=
1 |
4 |
1 |
n+1 |
n |
4n+4 |
又∵
1 |
an2 |
1 |
4n2 |
1 |
4n2?1 |
1 |
(2n?1)(2n+1) |
1 |
2 |
1 |
2n?1 |
1 |
2n+1 |
∴Tn=
1 |
a12 |