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## Main Question or Discussion Point

Is there a simple way to prove this? I tried to use induction, but you get down to some horrible fraction (letting [itex]N = 2m[/itex] for some [itex]m[/itex]) in the inductive step:

[tex]

\begin{pmatrix} 2(m+1) \\ k \end{pmatrix} = \frac{(2m)!}{k!(2m - k)!} \cdot \frac{(2m+2)(2m+1)}{(2m+2-k)(2m+1-k)}

[/tex]

The inductive hypothesis is to assume the thing on the left is biggest for [itex]k = m[/itex], but the second fraction gets bigger as you make [itex]k[/itex] bigger. So...what to do! Any comments? Thanks!

[tex]

\begin{pmatrix} 2(m+1) \\ k \end{pmatrix} = \frac{(2m)!}{k!(2m - k)!} \cdot \frac{(2m+2)(2m+1)}{(2m+2-k)(2m+1-k)}

[/tex]

The inductive hypothesis is to assume the thing on the left is biggest for [itex]k = m[/itex], but the second fraction gets bigger as you make [itex]k[/itex] bigger. So...what to do! Any comments? Thanks!