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Math Help - Triangle Numbers

  1. #1
    Member GAdams's Avatar
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    Triangle Numbers

    Triangle number sequence: 1,3,6,10,15,21,....

    nth term is given by 1/2n (n+1)

    a. give algebraic expression for the (n-1) term of the sequence.

    b. Prove that the sum of any two consecutive triangle numbers is a square number.
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  2. #2
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    Quote Originally Posted by GAdams View Post
    Triangle number sequence: 1,3,6,10,15,21,....

    nth term is given by 1/2n (n+1)

    a. give algebraic expression for the (n-1) term of the sequence.

    b. Prove that the sum of any two consecutive triangle numbers is a square number.

    Hello,

    to a): Plug in (n-1) instead of n into the given term:

    (1/2)(n-1)(n-1 + 1) = (1/2)n(n-1)


    to b)

    Add the nth term and the (n-1)th term:

    1/2n (n+1) + (1/2)n(n-1) = (1/2)n[(n+1) + (n-1)] = (1/2)n[2n] = nē
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  3. #3
    Member GAdams's Avatar
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    I don't understand how you cancelled down 'a' to get 1/2n(n-1). And why did you put brackets round the 1/2?

    Thanks
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  4. #4
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    Quote Originally Posted by GAdams View Post
    I don't understand how you cancelled down 'a' to get 1/2n(n-1).
    Hello,
    Code:
    (1/2)(n-1)(n    -1   +   1  ) = (1/2)(n-1) * n
                   \           /
                    equals zero

    Quote Originally Posted by GAdams View Post
    And why did you put brackets round the 1/2?

    Thanks
    As long as Latex isn't working writing mathematical expressions is a little bit tricky. Only to exclude any abiguity I use brackets to write a fraction. If you write
    Code:
                                              1
    1/2n(n-1) sometimes it is meant to be: --------
                                            2n(n-1)
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  5. #5
    Member GAdams's Avatar
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    Hmmm.

    I have it up to here:

    (1/2) (n-1) (n-1 + 1)

    You cancelled the -1 and +1 in the last brackets to get:

    (1/2) (n-1) *n

    Which to my 'logical' mind would multiply out to give:

    1/2n - 1/2 *n

    Which is zero! It seems that you have multiplied the 1/2 by the last n and left the brackets intact.

    How would I know to do that?
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  6. #6
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    Quote Originally Posted by GAdams View Post
    Hmmm.

    I have it up to here:

    (1/2) (n-1) (n-1 + 1)

    You cancelled the -1 and +1 in the last brackets to get:

    (1/2) (n-1) *n

    Which to my 'logical' mind would multiply out to give:

    1/2n - 1/2 *n

    Which is zero! It seems that you have multiplied the 1/2 by the last n and left the brackets intact.

    How would I know to do that?
    Hi,

    there are 2 points where I was obviously not exact enough:

    (1/2) (n-1) (n-1 + 1) = (1/2) (n-1) (n + 0) = (1/2) (n-1) (n)


    (1/2) (n-1) (n-1 + 1) = (1/2) (n-1) *n = 1/2nē - 1/2 *n . Because you multiply a sum you must use the law of distribution (not certain if this is the appropriate expression in English)
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