Divide the left and right sides of the above formula by 2 at the same time (n+1), and you get

a(n+ 1)/2^(n+ 1)=an/2^n+3*( 1/2)^(n+ 1)

Remember bn = an/2 n

∴b(n+ 1)=bn+3*( 1/2)^(n+ 1)

∴b(n+ 1)-bn=3*( 1/2)^(n+ 1)

∴bn-b(n- 1)=3*( 1/2)^n

……

b3-b2=3*( 1/2)^3

b2-b 1=3*( 1/2)^2

Superposition method, the left and right sides of the equation are all added.

bn-b 1=3*(( 1/2)^2+( 1/2)^3+……+( 1/2)^n)=3* 1/4*( 1-( 1/2)^(n- 1))/( 1- 1/2)=3/2*( 1-( 1/2)^(n- 1))

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

∫b 1 = a 1/2 = 5 * 2

∴bn=3/2-3*( 1/2)^n+b 1=4-3*( 1/2)^n

∴ an = bn * 2n = (4-3 * (1/2) n) * 2n = 2 (n+2)-3 is what you want.