By induction prove $\prod _{i=1}^{n} \frac{n+i}{2i-3} = 2^n (1 - 2n)$ I need to prove the following by induction.
$\forall n \in \Bbb N$
$\prod _{i=1}^{n} \frac{n+i}{2i-3} = 2^n (1 - 2n)$
I know the steps to take but I'm failing to come to the right side of the equation when $P(n+1)$. Could someone show me step by step how to do it so I can see where I'm making the mistake?
(There's a similar question already posted, but the solution doesn't help me identify my mistake)
 A: $$\prod_{i=1}^{n+1}\frac{n+1+i}{2i-3}=\prod_{i=1}^n\frac{n+i}{2i-3}\cdot\frac{(2n+1)(2n+2)}{(n+1)(2n-1)}\stackrel{\text{Inductive Hypotesis}}=$$$${}$$
$$=2^n\color{red}{(1-2n)}\frac{(2n+1)\cdot2\color{green}{(n+1)}}{\color{green}{(n+1)}\color{red}{(2n-1)}}=2^{n+1}\left(-1-2n\right)=2^{n+1}(1-2(n+1))$$
Explanation under request of the first equality above: Let us take first the denominator:
$$\prod_{i=1}^{n+1}(n+1+i)=(n+2)(n+3)\cdot\ldots\cdot(2n+2)$$
Thus, after the first equality, since we begin there with $\;n+1\;$ , we then divide by $\;n+1\;$ (since this factor does not appear on the left), and we also multiply by $(2n+1)(2n+2)\;$ since these factors don't appear in the product on the right.
About the denominators: it is
$$\prod_{i=1}^{n+1}\frac1{2i-3}=\frac1{-1\cdot1\cdot3\cdot\ldots\cdot(2n-1)}$$
whereas on the right side of first equality we don't have in that product the last factor $\;\frac1{2n-1}\;$ , and thus we multiply by it there 
A: Let $P_n=\prod_{i=1}^{n} \frac {n+1}{2i-3}$.
Consider the ratio:  $$\frac {P_{n+1}}{P_n}=\frac {2n+2}{2n-1}\;\prod_{i=1}^n \left(\frac {n+i+1}{2i-3}\times \frac {2i-3}{n+i}\right)=\frac {2n+2}{2n-1}\;\prod_{i=1}^n\frac {n+i+1}{n+i}$$  $$=\frac {2n+1}{n+1}\times \frac {2n+2}{2n-1}=2\times\frac {2n+1}{2n-1}$$
And the induction hypothesis now completes the proof.
