Question

trying to verify 1/(1*3*5*...*(2n-5)) = ((2^n)(2n-3)(2n-1)(n!))/(2n)! never seen a similar factorial problem

Answer 1

So are you supposed to prove it's an identity? Or establish that it is an identity?

Answer 2

I think the right side is supposed to reduce to look like the left side

Answer 3

What I mean is, do you have to show that the identity is true, or do you have to rewrite/simplify one side so that it matches the other exactly?

The first would involve a proof by induction, and the other could go any one of a number of ways.

Answer 4

unfortunately the only instructions I have on this problem is "Verify the formula: "

Answer 5

This should get you started:\[\begin{align*}
\frac{1}{1\times3\times5\times\cdots\times(2n-7)\times(2n-5)}&=\frac{2\times4\times\cdots\times(2n-6)\times(2n-4)}{(2n-5)!}\\
&=\frac{2(1)\times2(2)\times\cdots\times2(n-3)\times2(n-2)}{(2n-5)!}\\
&=\frac{2^{n-2}\bigg[1\times2\times\cdots\times(n-3)\times(n-2)\bigg]}{(2n-5)!}\\
&=\frac{2^{n-2}(n-2)!}{(2n-5)!}
\end{align*}\]
That's what you get just from working with the left side. To get what you're given on the right side, you would just multiply by whatever you need.

(made a small mistake)

Answer 6

Thanks! :-) I'm usually good at simplifying factorials, but this one has been a beast

Answer 7

yw