spamiam said:
Wow, with those 4 words of advice, I solved this in about a minute!
For another hint, Mentallic, remember the properties
<br />
\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}<br />
and
<br />
\binom{n}{k} = \binom{n}{n-k}.<br />
Oh wow I wish I remembered the first equality. For a question in an exam that doesn't provide us with something like "use the equality ..." but it instead prefers to tell me that tan=sin/cos, I feel like I just got slapped in the face.
\binom{2n}{n}=\binom{2n-1}{n}+\binom{2n-1}{n-1}
\binom{2n-1}{n-1}=\binom{2n-1}{2n-1-(n-1)}=\binom{2n-1}{n}
Therefore
\binom{2n}{n}=2\binom{2n-1}{n}
But all valid binomials are integers, thus it is even.
ehild said:
But it took me about half an hour to find these four words.
ehild
Some wise words indeed
As for the proof by induction, I had some trouble along the way, and I'm unsure of what 'excuse' to use.
Assume \binom{2n}{n}=2a, a\in Z
and now prove \binom{2n+2}{n+1}=2b, b\in Z
\binom{2n+2}{n+1}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)n!(n+1)n!}
=\frac{(2n+2)(2n+1)}{(n+1)^2}\binom{2n}{n}
Now all I can say at this point is that \binom{2n}{n} is some integer, so I need to show that \frac{(2n+2)(2n+1)}{(n+1)^2} is even for all integers n.
\frac{(2n+2)(2n+1)}{(n+1)^2}=2\left(\frac{2n+1}{n+1}\right)
And from this point I'm unsure, but I was also able to convert it into the form
=2\left(2-\frac{1}{n+1}\right) and from this expression, it's quite clear that it is going to not be true, so I must have done something wrong.
It doesn't really change anything though, because I still need to show that the other factor is an integer and it clearly isn't. It's obvious that I went wrong somewhere along the way, yet I can't see where.
