题目内容

(本小题满分12分)已知数列{an}的前n项和为Sn, 且满足条件:4S n =+ 4n – 1 , nÎN*.
(1) 证明:(a n– 2)2="0" (n ³ 2);(2) 满足条件的数列不惟一,试至少求出数列{an}的的3个不同的通项公式 .
(2) 当a1 =1且a n + an – 1 = 2时,得an ="1. " 2)当a1 =1且a n – a n – 1 =" 2" 时,得an =" 2n–1" .
3)当a1 =3且a n – a n – 1 =" 2" 时,得an =" 2n" + 1 .     4)当a1 =3且a n + an – 1 = 2时,得an =2(–1)n+ 1 + 1.
(1) 由条件4S n =+ 4n – 1 , nÎN*.得4S n – 1 =+ 4(n – 1 ) – 1,
相减得:4a n  =  + 4,化成–4a n+ 4–= 0,
∴ (a n– 2)2="0" .     4分
(2) 由(1)得:(a n –2 + an – 1 )(a n –2 – a n – 1 ) =" 0∴" a n + an – 1 =" 2 " 或a n – a n – 1 =" 2" . 2分
在4S n =+ 4n – 1中,令n = 1,得4a1 =+ 4 – 1,解得:a1 =1或 a1 ="3. " 2分
分四种情况:
1)当a1 =1且a n + an – 1 = 2时,得an =1.
2)当a1 =1且a n – a n – 1 =" 2" 时,得an =" 2n–1" .
3)当a1 =3且a n – a n – 1 =" 2" 时,得an =" 2n" + 1 .
4)当a1 =3且a n + an – 1 = 2时,得an =2(–1)n+ 1 + 1. 每个1分,有3个即可
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