(1) let b? n？ =a？ n？ -2n, verification: sequence {b? n？ } is a geometric series;

(2) Find the sequence {a? n？ {and the first n items of s? n？ .

Solution: b? n？ =a？ n？ -2n=3a？ n- 1？ -4n+6-2n=3a？ n- 1？ -6n+6=3a？ n- 1？ -6(n- 1)=3[a？ n- 1？ -2(n- 1)]=3b？ n- 1？

So b? n？ /b？ n- 1？ =3= constant, ∴{b? n？ } is the first item b? =a？ -2= 1-2=- 1, geometric series of q=3: b? n？ =-3? ;

Answer? n？ =b？ n？ +2n, the sum of the first n terms: s? n？ =∑b？ n？ +∑2n=-(3？ - 1)/2+(2+4+6+.....+2n)

=( 1-3? )/2+(2+2n)n/2=( 1-3？ )/2+n(n+ 1)

Note: ∑b? n？ Is the sum of the first n terms of geometric series; ∑2n is the sum of the first n terms of arithmetic progression, the first term is 2, and the tolerance is 2.