已知数列{an},{bn}满足a1=2,2an=1+anan+1,bn=an-1,设数列{bn}的前n项和为Sn,令Tn
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已知数列{an},{bn}满足a1=2,2an=1+anan+1,bn=an-1,设数列{bn}的前n项和为Sn,令Tn=S2n-Sn.
(Ⅰ)求数列{bn}的通项公式; (Ⅱ)判断Tn+1,Tn(n∈N*)的大小,并说明理由.
(Ⅰ)求数列{bn}的通项公式; (Ⅱ)判断Tn+1,Tn(n∈N*)的大小,并说明理由.
(Ⅰ)由bn=an-1得
an=bn+1代入2an=1+anan+1得2(bn+1)=1+(bn+1)(bn+1+1)
整理得bnbn+1+bn+1-bn=0
从而有
1
bn+1−
1
bn=1
∴b1=a1-1=2-1=1
∴{
1
bn}是首项为1,公差为1的等差数列,
∴
1
bn=n即bn=
1
n
(Ⅱ)Tn+1>Tn
证明:∵Sn=1+
1
2+
1
3+…+
1
n
∴Tn=S2n-Sn=
1
n+1+
1
n+2+…+
1
2n
Tn+1=
1
n+2+
1
n+3+…+
1
2n+
1
2n+1+
1
2n+2
Tn+1−Tn=
1
2n+1+
1
2n+2−
1
n+1
>
1
2n+2+
1
2n+2−
1
n+1=0
故Tn+1>Tn
an=bn+1代入2an=1+anan+1得2(bn+1)=1+(bn+1)(bn+1+1)
整理得bnbn+1+bn+1-bn=0
从而有
1
bn+1−
1
bn=1
∴b1=a1-1=2-1=1
∴{
1
bn}是首项为1,公差为1的等差数列,
∴
1
bn=n即bn=
1
n
(Ⅱ)Tn+1>Tn
证明:∵Sn=1+
1
2+
1
3+…+
1
n
∴Tn=S2n-Sn=
1
n+1+
1
n+2+…+
1
2n
Tn+1=
1
n+2+
1
n+3+…+
1
2n+
1
2n+1+
1
2n+2
Tn+1−Tn=
1
2n+1+
1
2n+2−
1
n+1
>
1
2n+2+
1
2n+2−
1
n+1=0
故Tn+1>Tn
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