How to Resolve Errors in Proof by Induction for a Factorial Ratio?

[Ryuky]

Homework Statement

So we have to prove that $\frac{(n+1)(n+2)(n+3)...(2n)}{1*3*5...*(2n-1)}$ = 2ⁿ2. 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 $\frac{(k+1)(k+2)(k+3)...(2k)}{1*3*5...*(2k-1)}$ = 2^k.

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

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

 

[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 divide 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