The maximum of $\binom{n}{x+1}-\binom{n}{x}$ The following question comes from an American Olympiad problem. The reason why I am posting it here is that, although it seems really easy, it allows for some different and really interesting solutions. Do you want to give a try?
Let $n$ be one million. Find the maximum value attained by $\binom{n}{x+1}-\binom{n}{x}$, where $0<x<n$ is an integer. 

Edit: I saw the answers below, and I can say that I posted this one because there is at least one really nice solution, which is not so "mechanical" :) The value of $n$ has no particular meaning, it stands just for a "sufficiently large integer"..

Note that
$$
\begin{align}
\binom{n}{x+1}-\binom{n}{x}
&=\left[\frac{n-x}{x+1}-1\right]\binom{n}{x}\\
&=\frac{n-2x-1}{x+1}\binom{n}{x}\tag{1}
\end{align}
$$
$(1)$ is positive for $x\lt\frac{n-1}2$ and negative for $x\gt\frac{n-1}2$.
Furthermore,
$$
\begin{align}
&\hphantom{}\left[\binom{n}{x+1}-\binom{n}{x}\right]-\left[\binom{n}{x}-\binom{n}{x-1}\right]\\
&=\left(\left[\frac{n-x}{x+1}-1\right]-\left[1-\frac{x}{n-x+1}\right]\right)\binom{n}{x}\\
&=\frac{4x^2-4nx+(n+1)(n-2)}{(x+1)(n-x+1)}\binom{n}{x}\tag{2}
\end{align}
$$
$(2)$ is $0$ when $x=\frac{n\pm\sqrt{n+2}}2$. If this $x$ is an integer, then $\binom{n}{x+1}-\binom{n}{x}=\binom{n}{x}-\binom{n}{x-1}$. Otherwise, the maximum occurs somewhere between $x$ and $x-1$.
Thus, the maximum value for $(1)$ is where $x=\left\lfloor\dfrac{n-\sqrt{n+2}}2\right\rfloor$.
For $n=1000000$, $x=499499$.
$$
\binom{1000000}{499500}-\binom{1000000}{499499}=9.582658295\times10^{301023}
$$
whereas
$$
\binom{1000000}{499499}-\binom{1000000}{499498}=9.582581749\times10^{301023}
$$
and
$$
\binom{1000000}{499501}-\binom{1000000}{499500}=9.582658257\times10^{301023}
$$
