已知數(shù)列{an},{bn}滿足a1=2,2an=1+anan+1,bn=an-1,設(shè)數(shù)列{bn}的前n項(xiàng)和為Sn,令Tn=S2n-Sn.
(Ⅰ)求數(shù)列{bn}的通項(xiàng)公式; (Ⅱ)判斷Tn+1,Tn(n∈N*)的大小,并說(shuō)明理由.
【答案】
分析:(I)將兩個(gè)已知等式結(jié)合得到關(guān)于數(shù)列{b
n}的項(xiàng)的遞推關(guān)系,構(gòu)造新數(shù)列,利用等差數(shù)列的通項(xiàng)公式求出
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,進(jìn)一步求出b
n.
(II)表示出T
n,T
n+1,求出T
n+1-T
n,通過(guò)放縮法,判斷出此差的符號(hào),判斷出T
n+1,T
n兩者的大�。�
解答:解:(Ⅰ)解:由b
n=a
n-1得
a
n=b
n+1代入2a
n=1+a
na
n+1得2(b
n+1)=1+(b
n+1)(b
n+1+1)
整理得b
nb
n+1+b
n+1-b
n=0
從而有
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∴b
1=a
1-1=2-1=1
∴
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是首項(xiàng)為1,公差為1的等差數(shù)列,
∴
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(Ⅱ)T
n+1>T
n證明:∵
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∴T
n=S
2n-S
n=
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故T
n+1>T
n點(diǎn)評(píng):求數(shù)列的通項(xiàng)公式時(shí),一般先看遞推關(guān)系的特點(diǎn)選擇合適的求通項(xiàng)的方法;求數(shù)列的前n項(xiàng)和一般也是先判斷通項(xiàng)的特點(diǎn),再選擇合適的方法.