How can I explain how I figured out the 99th Triangular Number?
I'd appreciate it if it is as simple as possible.
2 answers
Did you figure it out? What did you get? Are you referring to the series of numbers 1,3,6, 10, 14 etc?
You 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.
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.