Since C0 + C1 + C2 + C3 + …… + Cn = 2n
Putting n = 8, we get
C0+ C1 + C2 + C3 + …… + C8 = 28
But, sum of even coefficients = sum of odd coefficients
∴ C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7
Let C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = k
Now, C0 + C1 + C2 + C3 + C4 + C5 + C6 + C7 + C8 = 256
∴ (C0 + C2 + C4 + C6 + C8 ) + (C1 + C3 + C5 + C7 ) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C0 + C2 + C4 + C6 + C8 = C1 + C3 + C5 + C7 = 128