Let $n>1$ be an odd integer.
Show that there is an odd number of odd numbers in the sequence: $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}, ..., \binom{n}{\frac{n-1}{2}}$
I've tried doing it by separating the problem into two cases, one in which $n=4k+1$ and one in which $n=4k+3$ where $k$ is a full number. Then by supposing that $\binom{n}{i}$ is either even or odd for a given $i$, then by writing $\binom{n}{i+1}$ as $\frac{\binom{n}{i}}{i+1} (n-i)$ in order to determine the parity of $\binom{n}{i+1}$. I didn't get anywhere.
Could someone please help me? Thanks.
Ignoring $n=1$, the sum $\sum_{k=1}^{(n-1)/2} \binom nk$ is an odd sum:
$$\begin{align*} \sum_{k=0}^{n} \binom nk &= 2^n\\ \sum_{k=0}^{(n-1)/2} \binom nk &= 2^{n-1}\\ \sum_{k=1}^{(n-1)/2} \binom nk &= 2^{n-1} -1\\ &\equiv 1\pmod2 \end{align*}$$
If the contrary that there are even number of odd terms (plus any number of even terms), the sum would be even:
$$\begin{align*} \sum_{k=1}^{(n-1)/2} \binom nk &\equiv \left(\text{# of odd terms}\right)\cdot 1 + \left(\text{# of even terms}\right)\cdot 0 &&\pmod 2\\ \left(\text{# of odd terms}\right) &\equiv 1 &&\pmod2 \end{align*}$$