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?
 A: $$
\overbrace{\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2}} + \overbrace{\binom n {\frac{n+1} 2} + \cdots + \binom n {n-1} + \binom n n} = 2^n.
$$
The two sums under the $\overbrace{\text{overbraces}}$ are equal; hence
$$
\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2} = 2^{n-1}.
$$
Therefore
$$
\binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2} = 2^{n-1}-1 = \text{an odd number}.
$$
If a sum of finitely many terms is an odd number (as this one is) then the number of odd terms must be odd, since if it were even, then the sum would be even.
A: Using the fact that $\binom{n}{r}=\binom{n}{n-r}$, we can see that the sum of all the numbers on the sequence, call it $S$, satisfies $2S+2=2^n$.
Basically what we have done is consider the sum $T=\sum_{i=0}^n \binom{n}{i}$, which we know by the binomial theorem that $T=2^n$.
Now $\binom{n}{0}=1=\binom{n}{n}$. The sum of the remaining terms is $2S$. So we get $2S+2=2^{n}$.
This shows that $S=2^{n-1}-1$, and therefore $S$ is odd.
So $S$ must have an odd number of odd numbers in it.
A: 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.
