【答案】
分析:(1)方法一:直接根據條件求出b
n-1的表達式,再與b
n-2=的表達式作差,結合遞推關系式,整理即可證明數列{b
n}為等差數列;即可求出求其通項公式;
方法二:先根據數列{a
n}的遞推公式得到a
n+12=a
n+2-a
n+1+1;再代入b
n=a
12+a
22+…+a
n+22-a
1a
2…a
n+2整理可得b
n=n+3;即可說明結論.
(2)先求出c
n的表達式,進而得到
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=
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=
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=
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;再代入求出S
n,即可得到結論.
解答:解:(1)方法一 當n≥3時,因b
n-2=a
12+a
22+…+a
n2-a
1a
2…a
n①,
故b
n-1=a
12+a
22+…+a
n2+a
n+12-a
1a
2…a
na
n+1②. …(2分)
②-①,得 b
n-1-b
n-2=a
n+12-a
1a
2…a
n(a
n+1-1)=a
n+12-(a
n+1+1)(a
n+1-1)=1,為常數,
所以,數列{b
n}為等差數列. …(5分)
因 b
1=a
12+a
22+a
32-a
1a
2a
3=4,故 b
n=n+3. …(8分)
方法二 當n≥3時,a
1a
2…a
n=1+a
n+1,a
1a
2…a
na
n+1=1+a
n+2,
將上兩式相除并變形,得 a
n+12=a
n+2-a
n+1+1.…(2分)
于是,當n∈N*時,b
n=a
12+a
22+…+a
n+22-a
1a
2…a
n+2
=a
12+a
22+a
32+(a
5-a
4+1)+…+(a
n+3-a
n+2+1)-a
1a
2…a
n+2
=a
12+a
22+a
32+(a
n+3-a
4+n-1)-(1+a
n+3)
=10+n-a
4.
又a
4=a
1a
2a
3-1=7,故b
n=n+3(n∈N*).
所以數列{b
n}為等差數列,且b
n=n+3. …(8分)
(2)因 c
n=
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=
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,…(12分)
故
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=
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=
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=
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.
所以
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=
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,…(15分)
即 n<S
n<n+1. …(16分)
點評:本題綜合考查解決基本數列的基本方法(定義法,分組裂項求和等),考查運算能力.