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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