Originally posted by @wolfgang59From what I remember a triangular number is to do with the number of objects arranged in an equilateral triangle.
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
Originally posted by @vendaThe first is 1, then 3, 6, 10
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
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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?
Originally posted by @blood-on-the-tracksNice and neat.
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?
Originally posted by @vendaYou lost me.
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
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Originally posted by @wolfgang59It will be so, as we have a little pattern emerging, and so can prove the predicted formula is correct by induction.
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!
Originally posted by @wolfgang59Ok using this method, what is the sum of every number between 1 and 1000?
You lost me.
I could tell you within a few seconds
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Originally posted by @vendaYes, it is basically the first 'formula' in this sequence.
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!
Originally posted by @blood-on-the-tracksThanks.
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!.