1. Standard memberwolfgang59
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    16 Sep '17 13:10
    What is the sum of the first N triangular numbers?
    Show your working.
  2. Subscribervenda
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    16 Sep '17 15:29
    Originally posted by @wolfgang59
    What is the sum of the first N triangular numbers?
    Show your working.
    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
  3. Standard memberwolfgang59
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    16 Sep '17 21:26
    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
    The first is 1, then 3, 6, 10
    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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    17 Sep '17 10:552 edits
    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?
  5. Subscribervenda
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    17 Sep '17 12:08
    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
  6. Standard memberwolfgang59
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    18 Sep '17 08:31
    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?
    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?
  7. Standard memberwolfgang59
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    18 Sep '17 08:32
    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
    You lost me.
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    18 Sep '17 08:491 edit
    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 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 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!
  9. Subscribervenda
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    18 Sep '17 14:24
    Originally posted by @wolfgang59
    You lost me.
    Ok using this method, what is the sum of every number between 1 and 1000?
    I could tell you within a few seconds
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    18 Sep '17 14:38
    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
    Yes, it is basically the first 'formula' in this sequence.

    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!
  11. Standard memberwolfgang59
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    19 Sep '17 20:50
    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!
    Thanks.
    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!.
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