(本小题满分12分)已知数列{a n }的前n项和为S n , 且满足条件：4S n = + 4n – 1 , n&Ic

(2) 当a ₁ =1且a _n + a _{n – 1} = 2时，得a _n ="1. " 2）当a ₁ =1且a _n – a _{n – 1} =" 2" 时，得a _n =" 2n–1" .
3）当a ₁ =3且a _n – a _{n – 1} =" 2" 时，得a _n =" 2n" + 1 . 4）当a ₁ =3且a _n + a _{n – 1} = 2时，得a _n =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 + a _{n – 1} )(a _n –2 – a _{n – 1} ) =" 0∴" a _n + a _{n – 1} =" 2 " 或a _n – a _{n – 1} =" 2" . 2分
在4S _n = + 4n – 1中，令n = 1,得4a ₁ = + 4 – 1,解得：a ₁ =1或 a ₁ ="3. " 2分
分四种情况：
1）当a ₁ =1且a _n + a _{n – 1} = 2时，得a _n =1.
2）当a ₁ =1且a _n – a _{n – 1} =" 2" 时，得a _n =" 2n–1" .
3）当a ₁ =3且a _n – a _{n – 1} =" 2" 时，得a _n =" 2n" + 1 .
4）当a ₁ =3且a _n + a _{n – 1} = 2时，得a _n =2(–1) ^{n+ 1} + 1. 每个1分，有3个即可