Prove that if $0 \le k \le \frac {n-1}{2}$, then ${n \choose k} \le {n \choose k+1}$, with equality holding if and only if $k = \frac{n - 1}{2}$ Prove that if $0 \le k \le \frac {n-1}{2}$, then ${n \choose k} \le {n \choose k+1}$. Further, prove that equality is met if and only if $k = \frac {n-1}{2}$
I tried to use the contrapositive
$${n \choose k} > {n \choose k+1}$$
$$\frac {n!}{(n-k)! \cdot k!} > \frac {n!}{(n-k+1)! \cdot (k+1)!}$$
$$n! \cdot (n-k+1)! \cdot (k+1)! > n! \cdot (n-k)! \cdot (k)!$$
$$(n-k+1)(k+1) > 1$$
I'm trying to get to $k > \frac{n-1}{2}$ and I can't seem to get there. Can anyone help?
 A: \begin{align*}
\binom{n}{k + 1} - \binom{n}{k} & = \frac{n!}{(k + 1)![n - (k + 1)]!} - \frac{n!}{k!(n - k)!}\\
& = \frac{n!}{(k + 1)!(n - k - 1)!} - \frac{n!}{k!(n - k)!}\\
& = \frac{n!(n - k)}{(k + 1)!(n - k - 1)!(n - k)} - \frac{n!(k + 1)}{(k + 1)k!(n - k)!}\\
& = \frac{n!(n - k)}{(k + 1)!(n - k)!} - \frac{n!(k + 1)}{(k + 1)!(n - k)!}\\
& = \frac{n![n - k - (k + 1)]}{(k + 1)!(n - k)!}\\
& = \frac{n!(n - 2k - 1)}{(k + 1)!(n - k)!}
\end{align*}
For $0 \leq k \leq n$,
$$\frac{n!}{(k + 1)!(n - k)!} > 0$$
Hence, if $0 \leq k \leq n$,
$$\binom{n}{k + 1} - \binom{n}{k} = \frac{n!(n - 2k - 1)}{(k + 1)!(n - k)!}\geq 0 \iff n - 2k - 1 \geq 0 \iff \frac{n - 1}{2} \geq k$$
with equality holding if and only if $k = \dfrac{n - 1}{2}$.
A: $$\binom n{k+1}=\frac{n^\underline{k+1}}{(k+1)!}=\frac{n^\underline{k}(n-k)}{(k+1)k!}=\frac{n-k}{k+1}\cdot \frac{n^\underline{k}}{k!}=\frac{n-k}{k+1}\cdot \binom nk\\
\binom n{k+1}\ge\binom nk \;\;\;\text{ if }\frac{n-k}{k+1}\ge 1\;\;\;\Rightarrow n\ge 2k+1 \;\;\;\Rightarrow k\le\frac {n+1}2\qquad\blacksquare$$
