What could be a possible approach to find the proof of:
$\binom{2k+1}{k}$ is odd when $k=2^m-1$, otherwise $\binom{2k+1}{k}$ is even.
I have seen some similar problems in https://math.stackexchange.com/questions/317163 and https://math.stackexchange.com/questions/2046338, but I still don't know that why$\binom{2k+1}{k}$ is even when $k \neq 2^m-1$.
Any answer will be appreciated. Thanks!
Let $$k=2^{r_1}+2^{r_2}+\ldots+2^{r_n}$$ where $r_1,r_2,\ldots,r_n$ are nonnegative integers such that $r_1<r_2<\ldots<r_n$. Thus. $$2k+1=2^0+2^{r_1+1}+2^{r_2+1}+\ldots+2^{r_n+1}\,.$$ If there exists $j\in\{1,2,\ldots,n\}$ such that $r_j\neq r_{j-1}+1$ (here, $r_{0}:=-1$), then the bit corresponding to $2^{r_j}$ in $2k+1$ is $0$, whilst the bit corresponding to $2^{r_j}$ in $k$ is $1$, and $\displaystyle \binom{0}{1}=0$. By Lucas's Theorem, we conclude that $$\binom{2k+1}{k}\equiv 0\pmod{2}\,.$$ On the other hand, suppose that $r_j=r_{j-1}+1$ for $j=1,2,\ldots,n$. Then, $r_j=j-1$ for each $j=1,2,\ldots,n$, making $k=2^n-1$. By Lucas's Theorem, we get that $$\binom{2k+1}{k}\equiv\binom{1}{0}\cdot\Biggl(\binom{1}{1}\Biggr)^n=1\pmod{2}\,.$$
You could just use Legendre's formula or have some fun with the binomial expansion of $\left ( 1 + x \right)^{2k+1}$ modulo 2.
