Ans: $2^{\lceil n/2 \rceil}$
APPROACH:
for even length strings:
1st bit and nth bit are same. similarly, 2nd 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.
We can also write it as $2^{\lceil n/2 \rceil}$.