$$ \prod_{i=1}^n \left( 2i-1 \right) = \left( \frac{(2n)!}{n!2^{n}} \right) $$
My question for this proof is that I am trying to understand product notation and how to factor this question to its lowest form.
$\color{maroon}{Proof :}$
$\text{(i)}\; \; \text{The statement is true for } n = 1 $
$\left ( 2(1) -1) = \frac{2(1)!}{1!*2^{1}} \right) \Rightarrow 1 =1$
$\text{(ii)} \; \; \text{Let } n = k $
$$ \prod_{i=1}^k \left( 2i-1 \right) = \left( \frac{(2k)!}{k!2^{k}} \right) \text{Induction Hypothesis.}$$
$\text{Let } n= k+1 $
$$ \prod_{i=1}^{k+1} \left( 2i-1 \right) = \left( \frac{(2(k+1)!}{(k+1)!2^{k+1}} \right) $$
$ \left( 2(k+1)-1)* \frac{(2k)!}{(k)!2^{k}} \right) $ $\text{From my understanding this multiplies because it is product notation correct?}$
$\Rightarrow (2k+1)\frac{(2k)!}{k!2^{k}}$
This is where I get stuck because I cannot seem to reduce it anymore so that the right and left side equal each other. I am guessing I probably made a mistake with my arithmetic. Can anyone spot my mistake?
You've almost got the answer. Just multiply the numerator and denominator by $2k+2$. $$(2k+1)\frac{2k}{k!2^k}\times\frac{2k+2}{2k+2}$$