This is kind of a small math challenge, but the application is interesting.

First, you dont have to be a genius to figure this out i dont think, but you have to be a genius if you can figure out the application for this progression if you have never seen it before or have no idea where it came from before you started to solve it today.

Here is the progression:

2/sqrt(3), 8/sqrt(15), 18/sqrt(35), 32/(3*sqrt(7)), 50/(3*sqrt(11)), 72/sqrt(143)

Those are all EXACT not approximate, and the first is for n=1 and the last is for n=6, so they go for n's:

1,2,3,4,5,6

and each n has one of those values in that order.

Maybe i should write this out ... for each 'n' from 1 to 6:

[1] 2/sqrt(3)

[2] 8/sqrt(15)

[3] 18/sqrt(35)

[4] 32/(3*sqrt(7))

[5] 50/(3*sqrt(11))

[6] 72/sqrt(143)

The first goal is to figure out the progression in terms of 'n' so it will be a function of 'n'.

The second goal, much more difficult unless you've seen this before, is to figure out what application it came out of.

The first solution is interesting, the second one may be surprising if you've never seen this before.

Have fun