For example, if we take the case of Hexa-2.4-diene we have 3 isomers which will have cis-cis, trans-trans and cis-trans arrangements at the two double bonds present. The aim is to get a formula for the number of geometrical isomers of a symmetric polyene with n double bonds, when n is odd and when it is even.
I observed that cis-trans and trans-cis would mean the same thing for two double bonds at same position from the two sides. When I try to think what's going to happen after this it just gets too overwhelming because I think that one would have to consider a lot of cases.
I also read this post but it does not seem to help.
The answer is-
$2^{n-1}+2^{\frac{n-1}{2}}$ ; when $n$ is odd.
$2^{n-1}+2^{\frac{n}{2}-1}$ ; when $n$ is even.