2

I tried using induction, but in the inductive step, I get:

If $2n\choose n$ is divisible then I want to see that $2n +2\choose n +1$

$${2n +2\choose n +1} = (2n + 2)!/(n+1)!(n +1)! = {2n\choose n} (2n+1)(2n+2)/(n+1)^2$$

$$= {2n\choose n} 2(2n+1)/(n+1)$$

Then ${2n\choose n}$ is divisible by $2$ by hypothesis, but $2(2n+1)/(n+1)$ is not even an integer.

What can I do?

FranckN
  • 1,304
  • 3
    Hint: Pascal's triangle rule gives $${2n \choose n}={2n-1\choose n}+{2n-1\choose n-1}.$$ BTW, this is a duplicate. Will look for it. – Jyrki Lahtonen Feb 07 '14 at 14:42

2 Answers2

4

Try doing it directly rather than by induction - consider how many times 2 appears in the expansion of the numerator and denominator into prime factors.

meta
  • 1,025
1

Direct way: from this, corollary 14 in page 4:

Corollary 14. The power of prime $p$ dividing $n \choose k$ is the number of carries when you add $k$ to $n − k$ in base $p$ (and also the number of carries when you subtract $k$ from $n$ in base $p$)

So it is enough to proove that when you adding $n$ to itself in base 2 you have at least one carry, which is obvious follow from that $n$ have at least one 1 digit in binary.

LeeNeverGup
  • 2,671