# Show that there is an odd number of binomial coefficients when n is odd

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.

• Sorry, n>1. Lemme fix that. Commented Mar 5 at 2:02
• Ignoring $n=1$, show that $\sum_{k=1}^{(n-1)/2} \binom nk$ is an odd sum. If the contrary that there are even number of odd terms (plus any number of even terms), the sum would be even. Commented Mar 5 at 2:04
• Didn't think of that. Thank you very much! Commented Mar 5 at 2:17

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*}
\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*}