No doubts (B) is answer! how?
answers will be max( Y, X) we just need to find X in the terms of Y because we have two subsequences X and Y but X is not given inside MAX function in any options.
Suppose P = 20, 40, 32 hence Y = 20 + 40/2^1 + 32/2^2
= 20+20+8 = 48
Now we need to add a0 with our Y, a0 can be negative or positive because a is a real number can be -ve too. if a0 is -ve then
a0 + Y < Y // if a0 is -ve
lets calculate X now as a0 , a1/2^1, a2/2^2 ....... an-1/2^(n-1)
X = 30 + 20/2^1+ 40/2^2+ 32/2^3 = 30 + 10 + 10 + 4 = 54
X is 54, a0 is 30 and Y is 48
in the terms of Y, X will be = a0 + y/2 i.e. 30 + 48/2 = 54
hence MAX (Y, a0+ Y/2) is correct answer!