- 16 Sep '17 15:29

From what I remember a triangular number is to do with the number of objects arranged in an equilateral triangle.*Originally posted by @wolfgang59***What is the sum of the first N triangular numbers?**

Show your working.

Therefore the first triangular number would be 3(2, 1) the second 6(3,2,1) the third 10(4,3,2,1) and so on.

So the sum of the first 3 would be 19 - 16 Sep '17 21:26

The first is 1, then 3, 6, 10*Originally posted by @venda***From what I remember a triangular number is to do with the number of objects arranged in an equilateral triangle.**

Therefore the first triangular number would be 3(2, 1) the second 6(3,2,1) the third 10(4,3,2,1) and so on.

So the sum of the first 3 would be 19

1

1+2 = 3

1+2+3 = 6

1+2+3+4 = 10

etc.

the nth triangular number is n * (n+1) /2 - 17 Sep '17 10:55 / 2 editsThe formula for the sum of the first n triangular numbers is

n(n+1)(n+2)/6

Hence, to take the above, first 4 are 1, 3, 6, 10, clearly sum to 20

Put. n = 4 in formula, we get 4 x 5 x 6 all over 6 = 20

In proving it, are we allowing use of the formula for the sum of the first n square numbers and/ or natural numbers without proof of those? - 17 Sep '17 12:08It's similar to the quick method of finding the sum of any string of consecutive numbers.

For example the sum of 7,8,9,10,11

Let x = the first number

Let y = the last number

Let n = the number of intergers in the set

The formula is n (x + y )/2

So above example is 7 + 11 = 18

18/2 =9

9*5 = 45 - 18 Sep '17 08:31

Nice and neat.*Originally posted by @blood-on-the-tracks***The formula for the sum of the first n triangular numbers is**

n(n+1)(n+2)/6

Hence, to take the above, first 4 are 1, 3, 6, 10, clearly sum to 20

Put. n = 4 in formula, we get 4 x 5 x 6 all over 6 = 20

In proving it, are we allowing use of the formula for the sum of the first n square numbers and/ or natural numbers without proof of those?

I wonder what the sum from n=1 to n=N is for n(n+1)(n+2)/6 ???

An intuitive guess might be N(N+1)(N+2)(N+3)/24

Any idea? - 18 Sep '17 08:32

You lost me.*Originally posted by @venda***It's similar to the quick method of finding the sum of any string of consecutive numbers.**

For example the sum of 7,8,9,10,11

Let x = the first number

Let y = the last number

Let n = the number of intergers in the set

The formula is n (x + y )/2

So above example is 7 + 11 = 18

18/2 =9

9*5 = 45 - 18 Sep '17 08:49 / 1 edit

It will be so, as we have a little pattern emerging, and so can prove the predicted formula is correct by induction.*Originally posted by @wolfgang59***Nice and neat.**

I wonder what the sum from n=1 to n=N is for n(n+1)(n+2)/6 ???

An intuitive guess might be N(N+1)(N+2)(N+3)/24

Any idea?

It's a bit laborious to type out on here with all of the brackets and fractions, but does verify your predicted formula to be correct.

The sum of n triangular numbers is called a tetrahedron number...think of decreasing in sized triangles placed on top of each other (the 'sum' of triangular numbers) to form a tetrahedron.

Not sure where we would go with a similar name for sum of tetrahedron numbers! - 18 Sep '17 14:24

Ok using this method, what is the sum of every number between 1 and 1000?*Originally posted by @wolfgang59***You lost me.**

I could tell you within a few seconds - 18 Sep '17 14:38

Yes, it is basically the first 'formula' in this sequence.*Originally posted by @venda***Ok using this method, what is the sum of every number between 1 and 1000?**

I could tell you within a few seconds

The sum of integers from 1 to n is n(n+1)/2, which is basically what your explanation gives, jumping in at a number larger than 1.

I would add it is considerably easier to prove than either of the other 2 formulae mentioned on this thread! - 19 Sep '17 20:50

Thanks.*Originally posted by @blood-on-the-tracks***It will be so, as we have a little pattern emerging, and so can prove the predicted formula is correct by induction.**

It's a bit laborious to type out on here with all of the brackets and fractions, but does verify your predicted formula to be correct.

The sum of n triangular numbers is called a tetrahedron number...think of decreasing in s ...[text shortened]... a tetrahedron.

Not sure where we would go with a similar name for sum of tetrahedron numbers!

On the subject of higher dimensions I was recently watching a YouTube clip

on higher dimensional "spheres" contained in "cubes".

The spheres in higher dimensions are "bigger" than the "cubes" which contain them!.