Parity of an odd integer

Question

I'm not sure if I'm just tired and I'm missing something obvious, but how come I'm obtaining the following: $$(2m+1)^n=\sum_{k=0}^n \binom{n}{ k} (2m)^k = 2 \sum_{k=0}^n \binom{n}{ k} 2^{k-1}m^k $$ This seems to imply any power of an odd integer is even, but $3^2=9$ is an obvious counterexample.

Answer 1

When index $k=0$, $2^{k-1}=\frac12$, the corresponding term ,

$$2 \binom{n}{0}2^{0-1}m^0=1$$is odd.

Answer 2

All has been said:

$k=0$ spoils the party:

$\sum_{k=0}^{n} \binom{n}{k}(2m)^k=$

$ \binom{n}{0}(2m)^0 + 2\sum_{k=1}^{n}\binom{n}{k}(2)^{k-1}m^k.$

The first term in the above sum $=1.$

Hence?