F(n)=a 1+a2+..+ak, where k = 2 n.

f(n+ 1)=a 1+..+....+ak+a(k+ 1)+...+a(2k)

= f (n)+[A (k+1)+A (k+3)+...+A (2k-1)]+[A (k+2)+A (k+4)+... one (2k)]

= f (n)+[k+1+k+3 ...+2k-1]+[a (k/2+1)+a (k/2+2)+...+a (k)].

= f(n)+3k * k/4+[f(n)-f(n- 1)]

=2f(n)-f(n- 1)+3k^2/4

So f (n+1)-f (n) = f (n)-f (n-1)+3k2/4.

Namely:

f(3)-f(2)=f(2)-f( 1)+3/4*4^2

f(4)-f(3)=f(3)-f(2)+3/4*4^3

....

f(20 13)-f(20 12)=f(20 12)-f(20 1 1)+3/4*4^20 12

Add up the above categories:

f(20 13)-f(2)=f(20 12)-f( 1)+3/4*[4^2+4^3+..+4^20 12]

f(20 13)-f(20 12)= f(2)-f( 1)+3/4 * 4^2*(4^20 1 1- 1)/3=a3+a4+4^20 12-4=3+ 1+4^20 12-4=4^20 12

Choose C.