I'm confused as to why I'm getting these terms in this simple sequence incorrect.

Question

anonymous · Answer

attachment

directrix · Answer

Which problem are you doing? @JPeg16

anonymous · Answer

Weird, Sorry, I didn't realize that it sent the whole thing

anonymous · Answer

I'm on 5

anonymous · Answer

It's counting the last three answers wrong and I'm not sure why

directrix · Answer

On #5, I got the same answers as you did. Do you suppose that there is some special way in which the fractions of value less than 1 are supposed to be entered?

unklerhaukus · Answer

well im getting a different a_3

anonymous · Answer

maybe, I'll try that.  Glad to know you're getting the same answer though

unklerhaukus · Answer

$$a_n=\frac{(2n-1)!!}{n!}$$

$$a_3=\frac{(2\times3-1)!!}{3!}=5!!/3!=15/6$$

anonymous · Answer

Where did 5!! come from?

unklerhaukus · Answer

2x3-1=6-1=5

anonymous · Answer

ohhhh

anonymous · Answer

I see now

anonymous · Answer

So the next one would be 945/24?

directrix · Answer

That is the first time I have encountered the double factorial function. I took 5!! as (5!)! http://mathworld.wolfram.com/DoubleFactorial.html

unklerhaukus · Answer

$$a_4=\frac{(2\times4-1)!!}{3!}=\frac{7!!}{4!}=\frac{7\times5\times3\times1}{4\times3\times2\times1}$$

directrix · Answer

@UnkleRhaukus   Why is it that the first two terms were correct without realization that this is a double factorial problem.

anonymous · Answer

In the first two you would just multiply by 1

unklerhaukus · Answer

@Directrix

1!!=1!=1
2!!=2!=2

unklerhaukus · Answer

have you found a_4   @JPeg16  ?

anonymous · Answer

Yes, I got the rest

anonymous · Answer

Thank you!

unklerhaukus · Answer

good good