1^{st} bit and nth bit are same. similarly, 2^{nd} and (n-1)^{th} bits are same.
Now, 2 bits for every position, and we are filling n/2 positions. therefore $2^{n/2}$ strings

for odd-length strings

1st bit and nth bit are same. similarly, 2nd and (n-1)th bits are same. but {(n+1)/2}^{th} positions has 2 more ways to be filled
Now, 2 bits for every position, and we are filling n/2 positions. therefore $2^{n+1/2}$ strings.