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Math Help - Different of two consecutive triangulars is always a cube

  1. #1
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    Different of two consecutive triangulars is always a cube

    Show that the difference between two consecutive triangular numbres is always a cube.

    Proof. Let x be a triangular number, then x = k(k+1)/2 for some integers k.
    Let y be a consecutive triangular number, then y = (k+1)(k+2)/2.

    y-x = (k+2-k)(k+1)/2 = k+1

    But then there isn't any promise that it is a cube?
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  2. #2
    Senior Member JaneBennet's Avatar
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    You have stated the theorem incorrectly. The correct statement is:

    \color{white}.\quad. “The difference between the squares of two consecutive triangular numbers is always a cube.”

    Now try proving that.
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    thanks, I solved the problem now!
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  4. #4
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    Hello, tttcomrader!


    Please check the wording of that problem.

    The statement is patently impossible.


    \begin{array}{ccccccccccccccc}<br />
\Delta_n & 1 & & 3 & & 6 & & 10 & & 15 & & 21 & & \hdots\\<br />
\text{diff.} & & 2 & & 3 & & 4 & & 5 & & 6 & & \hdots\end{array}

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