Posted by PLEASE HELP on Wednesday, September 21, 2011 at 8:34pm.
Explicit formula for 8,15,29....

Algebra  tchrwill, Thursday, September 22, 2011 at 5:04pm
First glance flashes "finite difference series" but there is insufficient data to confirm same.
Assuming it is, the pattern would look like the following.
n.....1.....2.....3.....4.....5...
......8....15....29....50....78...
Diff....7.....14....21....28...
Diff.......7......7.....7....
With the second differences being constant, the nth terms are defined by a second order equation of the form a(n)^2 + b(n) + c = N
Using the given (assumed data)
a(1^2)+b(1^1)+c=8 or......a+b+c = 8
a(2^2)+b(2^1)+c=15.......4a+2b+c = 15
a(3^2)+b(3^1)+c(n)=29....9a+3b+c = 29
Subtracting each successive pair yields
3a + b = 7 and
5a + b = 14
Then, 2a = 7 making a = 3.5, b = 3.5 and c = 8
Therefore, any number in the sequence is defined by N = 3.5n^2  3.5n + 8
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