Explain equality $F(n)=2^{n-1} \cdot 1 \cdot 3 \dots(2n-3)=\frac{(2n-2)!}{(n-1)!}$

Question

$F(n)=2^{n-1} \cdot 1 \cdot 3 \dots(2n-3)=\frac{(2n-2)!}{(n-1)!}$

Any thouts how to prove this equality?

Thanks

Answer 1

\begin{align} 2^{n - 1} \cdot 1 \cdot 3 \cdot \ldots \cdot (2n - 5) (2n - 3) & = 2^{n - 1} \cdot \frac{1 \cdot 2 \cdot 3 \cdots \ldots \cdot (2n - 5)(2n - 4)(2n - 3)(2n - 2)}{2 \cdot 4 \cdot \ldots \cdot (2n - 4)(2n - 2)} \\ & = 2^{n - 1}\frac{(2n - 2)!}{2^{n - 1} \cdot 1 \cdot 2 \cdot \ldots \cdot (n - 2)(n - 1)} \\ & = \frac{(2n - 2)!}{(n - 1)!} \end{align}

Answer 2

\begin{align}1\cdot3\cdot5\cdots (2n-3) = \dfrac{(2n-2)!}{2(n-1)\cdot2(n-2)\cdot\cdots 2(1)} = \dfrac{(2n-2)!}{2^{n-1}(n-1)!} \end{align} Therefore, $$ 2^{n-1}\cdot1\cdot3\cdot5\cdots (2n-3)=\dfrac{(2n-2)!}{(n-1)!}$$

Answer 3

this is simple: $2^{n-1} \cdot 1\cdot3...(2n-3) \Rightarrow A = 1\cdot3...(2n-3) \Rightarrow A \times (2\cdot4\cdot6.....(2n)\cdot (2n-2)) = 1\cdot2\cdot3\cdot4\cdot5\cdot6...(2n-1)\cdot(2n-2)\cdot(2n-3) \Rightarrow A \times 2^{n-1}\cdot(n-1)!=(2n-3)! \Rightarrow 2^{n-1} \times A = \frac{(2n-3)!}{(n-1)!} $

Answer 4

Look first at a simple case, for instance $n=5$: you have $$ F(5)=2^4\cdot 1\cdot 3\cdot 5\cdot 7 $$ Insert the even numbers up to $8$, multiplying and dividing: $$ F(5)=2^4\,\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 7\cdot 8} {2\cdot 4\cdot 6\cdot 8} $$ The numerator is $8!=(2\cdot 5-2)!$ and the denominator can be rewritten as $2^4\cdot 1\cdot 2\cdot 3\cdot 4=2^4\cdot4!$, by collecting $2$ from each factor. The factors $2^4$ cancel with each other and you remain with $$ F(5)=\frac{8!}{4!} $$

Now the general case is the same, but maybe using induction is better: \begin{align} F(n+1) &=2^n\cdot1\cdot 3\cdot\ldots\cdot(2n-3)\cdot(2n-1) \\[6px] &=2\cdot (2n-1)\cdot F(n) \\[6px] &=2\cdot(2n-1)\cdot\frac{(2n-2)!}{(n-1)!} &&\text{induction hypothesis} \\[6px] &=2\cdot(2n-1)\cdot\frac{2n}{2n}\cdot\frac{(2n-2)!}{(n-1)!} &&\text{insert the even factors} \\[6px] &=2\frac{(2n-2)!\cdot(2n-1)\cdot(2n)}{(2n)\cdot(n-1)!} &&\text{rearrange the numerator} \\[6px] &=2\frac{(2n)!}{(2n)\cdot(n-1)!} &&\text{recognize the factorial} \\[6px] &=2\frac{(2n)!}{2(n\cdot(n-1)!)} &&\text{collect $2$ in the denominator} \\[6px] &=\frac{(2n)!}{n\cdot(n-1)!} &&\text{cancel $2$} \\[6px] &=\frac{(2n)!}{n!} &&\text{recognize the factorial} \\[6px] \end{align} The last expression is exactly what you get when you change $n$ into $n+1$ in $\frac{(2n-2)!}{(n-1)!}$.

Note that this is the same as Dominik's answer, but with all implicit passages worked out.