Parity of an odd integer 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.

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?
