Question

\[\frac{1}{2},\frac{3}{7},\frac{1}{3},\frac{5}{19},\frac{3}{14},...\] find \(a_{20}\)

Answer 1

best problem posted today
i miss foolformath and alchemy

Answer 2

@Loser66 you are not allowed to answer

Answer 3

why?

Answer 4

because you saw it already

Answer 5

I saw but I didn't understand, so???

Answer 6

oh ok

Answer 7

hehehe... just poking you

Answer 8

grrrr

Answer 9

I don't suppose it's as simple as \(a_n=a_{n-4}a_{n-3}\)?

Answer 10

no i don't think so

Answer 11

no, definitely not

Answer 12

7/52 i cant read what it wants you to find

Answer 13

no no 7/39

Answer 14

you've missed your shot.

Answer 15

twice

Answer 16

\[a_n = (n+1)/(n^2 + 3)\]

Answer 17

not sure who you are saying but yup... that is what i did... which is how i got a6=7/39

Answer 18

a_20 = (20+1)/[(20^2)+3]

Answer 19

maybe
\[a_{20} = (20+1)/[(20^2)+3]\]

Answer 20

or \[\frac{21}{4003}\] if you prefer

Answer 21

someone just asked this question, thought it was a good one
time to bring some more math to the math portion of open study
not that i mind finding the slope of a line mind you...

Answer 22

let me try a_6
a_6 = (6+1)/[(6^2)+3]
a_6 = 7/39

Answer 23

I have math questions...But people don't don't understand them :(

Answer 24

let me try if that works with 4th term
a_4 = (4+1)/[(4^2)+3]
a_4 = 5/19

Answer 25

@KingGeorge may I ask if there's any simpler way for vector addition

Answer 26

probably myself included!

Answer 27

remember alchemy? wild questions

Answer 28

I remember alchemy from 3rd world country
performed by quack doctors
-.-

Answer 29

Vector addition? That's pretty darn simple in the first place.

Alchemy was mostly before my time I think. Or I was just never on at the same time.

Answer 30

back in the good old wild west days of openstudy, before all the rules and annoying children spamming the chats
used to crash every night at around 8 or so

Answer 31

I've been here only for about a year, and I am still learning basic math