I read that a group of perfect Hamming code is defined as $[2^r-1,2^r-r-1,3]$ for any integer $r\geq 2$.
To be perfect such codes should satisfy the following equivalence (derived from the Hamming bound): $$\sum_{j=0}^r\binom{2^r - 1}{j} = 2^{r}$$ which is in contradiction with the fact that $r\geq2$.
What am I getting wrong?
A linear $[n,k,d]_q$-code is perfect if and only if $$ q^{n-k}=\sum_{i=0}^{(d-1)/2}\binom{n}{i}(q-1)^i. $$ In the case of a binary Hamming code, $n=2^r-1$, $k=2^r-r-1$ and $d=3$, $q=2$. So this gives $$ 2^r=\sum_{i=0}^1\binom{2^r-1}{i}. $$ Do you see your mistake?