Prove by induction that $\displaystyle\prod_{i=1}^{n}\frac{n+i}{2i-3}=2^n(1-2n)$. Prove by induction that $\displaystyle\prod_{i=1}^{n}\frac{n+i}{2i-3}=2^n(1-2n)$.
What I did:
It's easy to see that the base case is true. Then I have to prove that if the statement is true for $n$ then it must be true for $n+1$. That is to say, if true for $n$, prove that $\displaystyle\prod_{i=1}^{n+1}\frac{n+1+i}{2i-3}=2^{n+1}(1-2(n+1))=\left(\prod_{i=1}^{n}\frac{n+1+i}{2i-3}\right)\frac{2n+2}{2n-1}=-2\cdot2^n(2n+1)$. Of course the $+1$ inside the product is blocking my progress, because I can't replace it by the inductive hypothesis. I've figured that $-2\cdot2^n(2n+1)=-2\cdot2^n(1-2n)-n2^{n+3}$, but I don't know if that helps.
Help is much appreciated!
 A: Brute-force induction won't work well here, it's better to approach this problem by looking for simplifications. For example, the numerator in the product can be written in the form $\frac{(2n)!}{n!}$ and can be removed from the product entirely, thus leaving $\displaystyle\frac{(2n)!}{n!}\prod_{i=1}^{n}\frac{1}{2i-3}=2^n(1-2n)$. The same reasoning should be used for the denominator. It is $\frac{1}{(-1) \cdot (1) \cdot (3) ... \cdot(2n -3)}$
which is also $-\frac{2^{n-1}(n-1)!}{(2n-2)!}$, so you have $-\frac{2^{n-1}(n-1)!}{(2n-2)!}\frac{(2n)!}{n!} = 2^{n} (1-2n) \implies \frac{2^{n-1}(n-1)!}{(2n-2)!}\frac{(2n)!}{n!} = 2^{n} (2n-1)$.
Further simplifications show that $\frac{2n(2n-1)}{n} = 2(2n-1)$ which is obviously an equality. Thus the statements are equal. You shouldn't use induction here.
A: Since induction is required, consider this:
$\displaystyle\prod_{i=1}^{n+1}\frac{n+1+i}{2i-3}=\prod_{i=1}^{n}\frac{n+i}{2i-3}\Big(\frac{(2n+1)(2n+2)}{(2n-1)(n+1)}\Big) \implies$
$2^{n+1}(2n+1) =  2^n(2n-1) \frac{(2n+1)(2n+2)}{(2n-1)(n+1)}$ is true under assumption, which means assuming the first case is true, induction holds.
A: The inductive hypothesis is:
$$\prod_{i=1}^{k} \frac {k+i}{2i-3}=2^k(1-2k)$$
LHS can be written as:
$$\frac {k+1}{-1} \cdot \frac {k+2}{1} \cdot \frac {k+3}{3}...\cdot \frac {k+k}{2k-3}=\left(\frac {k+2}{-1} \cdot \frac {k+3}{1} \cdot \frac {k+4}{3}...\cdot \frac {k+k}{2k-5}\right)\cdot \left(\frac {k+1}{2k-3}\right)=\left(\left(\prod_{i=1}^{k+1} \frac {k+1+i}{2i-3}\right)\left(\frac {2k-3}{2k+1} \cdot \frac {2k-1}{2k+2}\right)\right) \cdot \left(\frac {k+1}{2k-3}\right)=2^k(1-2k)$$
Now simplify the terms other than the $\prod$.
You are left with:
$$\prod_{i=0}^{k+1} \frac {k+1+i}{2i-3}=-2^{k+1}(1+2k)$$
This completes the inductive step.
