Let n= 1 get 2=a 1+2a 1.

a 1=2/3

Go n+ 1

2=a(n+ 1)+2s(n+ 1)

Subtract these two expressions.

0 = 2-2 = a(n+ 1)-a(n)+2[s(n+ 1)-s(n)]= a(n+ 1)-a(n)+2a(n+ 1)= 3a(n+ 1)-a(n)，

a(n+ 1)=( 1/3)a(n)，

a(n+ 1)/a(n)= 1/3

So {a(n)} is the geometric series of the first term a( 1)=2/3 and the common ratio (1/3).

a(n)=(2/3)( 1/3)^(n- 1)= 2/3^n