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This was in one of the examples of the textbook, but I couldn't figure out how they solved it. They say they multiply the left hand side by $\frac{n!}{n!}$ to get the right hand side: $$ \frac{2^n \cdot (2n-3)!!}{n!} = 2\frac{(2n-2)!}{n!(n-1)!} $$ The double factorial stands for the product of all odd integers from 1 to 2$n$-3. I've given this problem a lot of time, and I'm all out of ideas at this point. |
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Notice that $$2^n n! = 2(n) \cdot 2(n-1) \cdot 2(n-2) \cdots 2(2) \cdot 2(1) = (2n)(2n-2) \cdots (4)(2)$$ is the product of all even integers between $1$ and $2n$. So when you multiply the numerator and denominator by $n!$ you end up with $$\dfrac{2n(2n-2)!}{n!n!}$$ since the even terms between $1$ and $2n-2$ interweave between the odd terms. Cancelling the rogue $n$ on the numerator gives the result you seek. |
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All you need is: $$ K:=2^{n-1}(n-1)!=(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdots (2\cdot(n-1))=2\cdot 4\cdots (2n-2). $$ This gives for $K$ the product of all even integers between $1$ and $2n-2$. Therefore: $$(2n-2)!/K=(2n-3)!!$$ And this why your LHS turns into your RHS. |
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