Can we prove by induction that for every integer $k >1 $ , $2+2^k \choose r$ is even for all $2 < r \le n=1+2^{k-1}$ ? Or by some divisibility properties of Binomial co-efficients ?
I wanted to start as $2^k \choose r$ $+$ $2$$2^k \choose r-1$ $+$ $2^k \choose r-2$ $=$$ 2+2^k \choose r$ but I don't know whether $2|{2^k \choose m}$ or not ... Please help . Thanks in advance
After having written
$$\binom{2+2^k}{r} = \binom{2^k}{r} + 2\binom{2^k}{r-1} + \binom{2^k}{r-2},$$
it clearly suffices to see that
$$\binom{2^k}{m} \equiv 0 \pmod{2}$$
for $1 \leqslant m \leqslant 2^{k-1}$. Now we can write
$$\binom{2^k}{m} = \frac{(2^k)!}{m!(2^k-m)!} = \frac{2^k}{m}\binom{2^k-1}{m-1},$$
and to conclude that $\binom{2^k}{m}$ is even, we only need to see that the numerator of $\frac{2^k}{m}$ in the reduced form is even. Writing $m = 2^\mu\cdot \rho$ with odd $\rho$, it follows that
$$\binom{2^k}{m} = 2^{k-\mu}\Biggl[\frac{1}{\rho}\binom{2^k-1}{m-1}\Biggr]$$
is even since $m \leqslant 2^{k-1}$ implies $\mu < k$.