math
posted by paige .
Write an explicit formula for the sequence onehalf, threesevenths, onethird, fivenineteenths, threefourteenths, ... Then find a14. (1 point)
an = an – 1 – n minus one over seven n; fifteeen over one hundred nintey nine
an = a sub n plus one over n squared plus three ; fifteen over one hundred nintey nine
an = n plus one over n squared plus three ; fifteen over one hundred nintey nine
an = n over n cubed minus one ; fifteen over one hundred nintey nine

what's with all the words? We have perfectly good digit keys...
1/2, 3/7, 1/3, 5/19, 3/14, ...
Looks pretty random, but we know we are after something we can easily relate to n. What if we get the denominators steadily increasing? Rewrite the sequence as
2/4, 3/7, 4/12, 5/19, 6/28
So, now it's easy to see that
an = (n+1)/(n^2+3)
a11 = 12/124 = 3/31
Dunno quite what to make of all the verbage below that. Try using some symbols and asking a question. 
Hahaha! Oh Steve you crack me up with all that verbage and question asking. Anyways the question said to find a14, NOT a11. a14 = 15/199. So out of the wordy answer choices Paige gave, the answers C.
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