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September 20, 2014

September 20, 2014

Posted by **Bob** on Tuesday, January 22, 2008 at 8:21pm.

I'd appreciate it if it is as simple as possible.

- Math -
**drwls**, Tuesday, January 22, 2008 at 8:45pmDid you figure it out? What did you get? Are you referring to the series of numbers 1,3,6, 10, 14 etc?

- Math -
**tchrwill**, Wednesday, January 23, 2008 at 10:41amYou might examine the initial sequence of triangular numbers to if they had some property that would allow you to derive the nth triangular number?

n.....1.....2.....3.....4.....5......6...

N...1.....3.....6....10...15....21...

DIff...2.....3.....4......5.....6

Diff......1......1......1.....1

Since the second differences are equal, the sequence is a finite difference sequence making the nth term of the form an^2 + bn + c = 0

Using the data derived:

1--a(1)^2 + b(1) + c = 1 or a + b + c = 1

2--a(2)^2 + b(2) + c = 3 or 4a + 2b + c = 3

3--a(3)^2 + b(3) + c = 6 or 9a + 3b + c = 6

4--Subtracting (2) from (1) yields 3a + b = 2

5--Subtracting (3) from (2) yields 5a + b = 3

6--Subtracting (4) from (5) yields 2a = 1 making a = 1/2

7--Substituting a = 1/2 back into (2) yields b = 1/2

8--From (1), c = 0

Therefore, the nth term, Nn = n^2/2 + n/2 + 0 = (n^2 + n)/2 = n(n + 1)/2.

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