Solution
= Solution
Adding the all-one codeword is a bijection from $C$ to itself, and it replaces every bit by its complement. It therefore sends weight $j$ to weight $n-j$, proving
$$
\nu(n-j)=\nu(j).
$$
The code has one word of weight zero and, since its <minimum Hamming distance of a linear code>[minimum Hamming distance] is three, no words of weight one or two. Combining this with part (i) gives
$$
\boxed{\nu(n-1)=0,\qquad
\nu(n-2)=0,\qquad
\nu(n-3)=\frac{n(n-1)}6}.
$$