<s>[INST] 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?
 [/INST] $$\binom n{k+1}=\frac{n^\underline{k+1}}{(k+1)!}=\frac{n^\underline{k}(n-k)}{(k+1)k!}=\frac{n-k}{k+1}\cdot \frac{n^\underline{k}}{k!}=\frac{n-k}{k+1}\cdot \binom nk\\
\binom n{k+1}\ge\binom nk \;\;\;\text{ if }\frac{n-k}{k+1}\ge 1\;\;\;\Rightarrow n\ge 2k+1 \;\;\;\Rightarrow k\le\frac {n+1}2\qquad\blacksquare$$
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