an=f(2^n)=2a(n- 1)+2^(n- 1)2=2a(n- 1)+2^n

That is, an = 2a (n- 1)+2 n.

Divide both sides by 2 n at the same time

an/2^n=a(n- 1)/2^(n- 1)+ 1

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

Let the new series {an/2 n} be bn.

Then bn is arithmetic progression, and b 1=a 1/2= 1.

bn= 1+(n- 1)=n

an=n2^n