Help with sequence formula

  • Thread starter Karla
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Hi all,

Im doing some work with the triangular numbers, pretty basic stuff:

1 3 6 10 15

Now im trying to understand how to get the formula for calculating the nth term +1, the formula is:

n(n+1)/2

I have tried doing a difference table, only to find that the second difference is 1, which lead me to think the forumla must start with 2N but this is incorrect. Please can someone show me the easiest method for working out basic sequence formulas.

Thanks,
Karla
 
Last edited:

HallsofIvy

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Karla said:
Hi all,

Im doing some work with the triangular numbers, pretty basic stuff:

1 3 6 10 15

Now im trying to understand how to get the formula for calculating the nth term +1, the formula is:

n(n+1)/2

I have tried doing a difference table, only to find that the second difference is 1, which lead me to think the forumla must start with 2N but this is incorrect. Please can someone show me the easiest method for working out basic sequence formulas.

Thanks,
Karla
It's not clear to me why the fact that the second difference is 1 would lead you to think that the formula must start with 2N!
Newton's divided difference formula says that if f(1)= a0, the first difference (at 1) is a1, second difference a2, third difference a3, etc. Then
[tex] f(n)= a_0+ a_1(n-1)+ \frac{a_2}{2}(n-1)(n-2)+ \frac{a_3}{3!}(n-1)(n-2)(n-3)+ ...[/tex].

If the second difference is a constant then all succeeding differences are 0 and the formula gives a polynomial. In particular, for the sequence of triangular numbers, the first differences are just the sequence of counting numbers and the second difference is 1 for all n. At n= 1, the value is 1, the first difference is 2 and the second difference is 1 with all succeeding differences 0. Newton's divided difference formula gives
[tex]1+ 2(n-1)+ \frac{1}{2}(n-1)(n-2)[/tex]
[tex]= 1+ 2n- 2+ \frac{1}{2}(n^2- 3n+ 2)[/tex]
[tex]= 2n+ \frac{1}{2}n^2- \frac{3}{2}n[/tex]
[tex]= \frac{1}{2}n^2+ \frac{1}{2}n[/tex]
[tex]= \frac{n(n+1)}{2}[/tex]
 
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There is a great way to work out that formula for the first n numbers. This method was supposedly used by Gauss when he was 10. His teacher, wanting to leave the classroom for a time, told the students to add up the first 100 numbers and get the total. (This is the 100th triangle number.) But, before the teacher could get out of the room, Gauss presented the answer!

His method: Consider the series 1,2,3,......100. Now reverse this on the next line: 100,99,98......1. and add the two series together term by term. The result is 101 written 100 times, and the correct answer is 1/2 of that! which is 50x101 = 5050. http://www.cut-the-knot.org/Curriculum/Algebra/GaussSummation.shtml
 
Last edited:
Here is what I found to be a nice way of finding an equation for a sequence assuming it is a sequence and not just a random set of numbers. I had it as a power point but had to save it as a PDF file to attach it. Let me know if it helps.
Jim

PS: For your example.......1 3 6 10 15
3-1=2 2 3 4 5
6-3=3 1 1 1
10-6=4
15-10=5
3-2=1
4-3=1
5-4=1
The '1' repeats at the second row so we start with T=1/(2!) n^2
And 1/2 n^2 has a sequence of 1/2 2 9/2 8.
We take our sequence and subtract these to get a remainder sequence and then repeat the first part again.
1/2 1 3/1 2
1/2 1/2 1/2
This one started repeating in the first row so we add that to the first part we found and get
T=1/2 n^2 + 1/2 n and have no remainder this time so we are done.

I explained these steps a lot more thorough in the PDF file.
 

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HallsofIvy

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Yes, that is precisely the "Newton's divided difference formula" that both I and Robert Ihnot referred to five years ago!
 

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