Prove that the sequence of combinations contains an odd number of odd numbers 
Let $n$ be an odd integer more than one. Prove that the sequence $$\binom{n}{1}, \binom{n}{2}, \ldots,\binom{n}{\frac{n-1}{2}}$$
  contains an odd number of odd numbers.

I tried writing out the combination form as $$\frac{(2k+1)!}{(m!)((2k+1)-m)!}.$$ How do I use this to show that the sequence contains an odd number of odd numbers?

Suppose $n=2k+1$
Note that $\binom{2k+1}{1}=\binom{2k+1}{2k}$, $\binom{2k+1}{2}=\binom{2k+1}{2k-1}$,...,$\binom{2k+1}{k}=\binom{2k+1}{k+1}$.
Thus 
$$\binom{2k+1}{1}+\binom{2k+1}{2}+\cdots+\binom{2k+1}{k}+\binom{2k+1}{k+1}+\cdots+\binom{2k+1}{2k-1}+\binom{2k+1}{2k}=2^{2k+1}-2$$
From the above considerations, 
$$2\binom{2k+1}{1}+2\binom{2k+1}{2}+\cdots+2\binom{2k+1}{k}=2^{2k+1}-2.$$ Therefore
$$\binom{2k+1}{1}+\binom{2k+1}{2}+\cdots+\binom{2k+1}{k}=2^{2k}-1 \tag{$*$}$$
If the number of odd terms in $\binom{2k+1}{1},\binom{2k+1}{2}, \ldots, \binom{2k+1}{k}$ is even, then the sum in $(*)$ is even. This contradiction shows that the number of odd terms is odd.
