If f (3) = 4xlog2 (3) + 233, then the value of F (2) + F (4) + F (8) +... + F (2) is?

If f (3) = 4xlog2 (3) + 233, then the value of F (2) + F (4) + F (8) +... + F (2) is?

F (3) = 4xlog2 (3) + 233
=4log2(3^x)+233
So f (x) = 4log2x + 233
Bring in F (2) + F (4) +,... + F (2 ^ 8)
=4*(1+2+3+4+5+6+7+8)+233*8
=4*9*8/2+233*8
=144+1864
=2008