Is it true that if $n = 2^k$ for some integer $k\geq 1,$ then $\binom{n}{\frac{n}{2}}$ is an even number?

Question

Question: Fix a natural number $n.$ Is it true that if $n = 2^k$ for some integer $k\geq 1,$ then $$\binom{n}{\frac{n}{2}}$$ is an even number?

Note that $\frac{n}{2}$ is always an integer, so the combination is defined.

Recall that to calculate $\binom{n}{\frac{n}{2}},$ we calculation a fraction with numerator $\frac{n}{2}$ numbers starting with $n$ and then decrease by $1,$ and denominator $\frac{n}{2}$ starting with $1$ and then increase by $1.$

I notice that $n$ in numerator always cancels with $2 \times \frac{n}{2}$ at the denominator, so the number of even numbers for numerator is more than in denominator. However, I do not take into account the power of $2$ in both numerator and denominator.

Any hint would be appreciated.

$$2^{2n}={2n\choose n}+2\sum_{k=0}^{n-1}{2n\choose k}$$ – Did Mar 08 '18 at 11:20 — Did, Mar 08 '18 at 11:20

score 2 · Answer 1 · answered Mar 08 '18 at 11:51

2

For any integer $m\ge1$, $$\binom{2m}{m}=\binom{2m-1}{m-1}+\binom{2m-1}m=2\binom{2m-1}{m-1}$$ is even.

answered Mar 08 '18 at 11:51

Angina Seng

158,341

Is it true that if $n = 2^k$ for some integer $k\geq 1,$ then $\binom{n}{\frac{n}{2}}$ is an even number?

1 Answers1