Number of binary codes with parameters $(n, 2, n)$ for $n ≥ 2$ !? I did the problem with my knowledge, I don't know whether it is True or false
For $n\geq 2$
$\{00000(n terms),11111(n terms)\}$ is the trivial one
Consider the first code if we change the first coordinates of first code then we must change the first coordinates of 2nd code. Through changing 1 coordinate. So we get such $nC1$ number of codes with the parameter.
by changing 2 coordinates we get $nC2$
we only need to find up to $\lfloor n/2 \rfloor$ since if we find for $i$ changing coordinates , or $n-i$ are same
So for each $n\geq 2$ There are $\sum _{i=0}^{\lfloor n/2 \rfloor}nCi$ code word with parameter $(n, 2,n)$
Your answer is correct for odd $n$, but note that the expression can be simplified to $2^{n-1}$. For even $n$, you produce each code twice in the $i = n/2$ case (as both codewords can be found by changing $n/2$ coordinates of the zero word). Following your approach, the correct expression in the $n$ even case is $\sum_{i=0}^{n/2-1} \binom{n}{i} + \frac{1}{2}\binom{n}{n/2}$, which again equals $2^{n-1}$.
A simpler reasoning is the following:
The first codeword is arbitrary, so $2^n$ possibilities. The second one is then fixed as the complement of the first. In this way, each code is constructed twice, by interchanging the two codewords. So the number of these codes is $2^{n-1}$.