# Proof by Induction Problem

1. Oct 15, 2011

### Ryuky

1. The problem statement, all variables and given/known data

So we have to prove that $\frac{(n+1)(n+2)(n+3)...(2n)}{1*3*5...*(2n-1)}$ = 2n

2. The attempt at a solution

I. For n=1, obviously the proposition is true. (2*1/(2-1) = 2^1 = 2)

II. Let n=k and assume $\frac{(k+1)(k+2)(k+3)...(2k)}{1*3*5...*(2k-1)}$ = 2k.

Now, for n=k+1 we have: 2k * $\frac{(2k +2)}{2k+1)}$ = 2k+1 → 2(k+1)*$\frac{(k+1)}{2k+1)}$ = 2k+1. Which is not true.

So, I cannot figure out what am I doing wrong.
Thanks in advance...

2. Oct 15, 2011

### susskind_leon

You forgot that in the nominator, you're starting at n+1. So, by doing the transition k->k+1, you need to devide by k+1 also to take that into account. Also, you need to multiply by (2k+1)*(2k+2), since in the nominator you have the product over all numbers between k+1 and 2k.
So, what you get is: (2k+1)*(2k+2)/(k+1)/(2k+1)=2

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook