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Math Help - Prove: each square is the sum of 2 consecutive triangular numbers

  1. #1
    Member courteous's Avatar
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    Question Prove: each square is the sum of 2 consecutive triangular numbers

    Is the following an acceptable proof (be harsh )?

    t_k=1+2+3+\ldots+k=\frac{(k+1)k}{2}
    t_{k-1}=1+2+3+\ldots+(k-1)=\frac{k(k-1)}{2}
    \Longrightarrow t_{k-1}+t_k=k^2 \text{ } \square ?

    I mean, is the induction also required? Even if not, I need help with it:

    • Base case - k=1: 1^2=t_0+t_1=0+1=1; define t_0=0.


    • Induction - k=k: suppose k^2=t_{k-1}+t_k.
      Then (k+1)^2=t_k+t_{k+1}=? ... how do I use the supposition?
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  2. #2
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    Just show that

    \frac{(k + 1)k}{2} + \frac{(k - 1)k}{2} = k^2...


    \frac{(k + 1)k}{2} + \frac{(k - 1)k}{2} = \frac{k^2 + k}{2} + \frac{k^2 - k}{2}

     = \frac{k^2 + k + k^2 - k}{2}

     = \frac{2k^2}{2}

     = k^2.
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  3. #3
    Member courteous's Avatar
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    Done that, just didn't write it down. The Halmos \square sign (with a "?" mark) indicates whether this is enough ... or is induction also needed (even if it is not, still help me with the induction step ).

    MOD request: Can you please restore my spam-hijacked question?
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    t_{k + 1} + t_k = \frac{(k + 2)(k + 1)}{2} + \frac{(k+1)k}{2}

     = \frac{k^2 + 3k + 2}{2} + \frac{k^2 + k}{2}

     = \frac{2k^2 + 4k + 2}{2}

     = \frac{2(k^2 + 2k + 1)}{2}

     = k^2 + 2k + 1

     = (k + 1)^2.
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  5. #5
    Member courteous's Avatar
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    Is this induction (not saying that it isn't, I just don't know)? I mean, isn't it necessary to use k^2=t_{k-1}+t_k equality?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by courteous View Post
    Is the following an acceptable proof (be harsh )?

    t_k=1+2+3+\ldots+k=\frac{(k+1)k}{2}
    t_{k-1}=1+2+3+\ldots+(k-1)=\frac{k(k-1)}{2}
    \Longrightarrow t_{k-1}+t_k=k^2 \text{ } \square ?

    I mean, is the induction also required? Even if not, I need help with it:

    • Base case - k=1: 1^2=t_0+t_1=0+1=1; define t_0=0.


    • Induction - k=k: suppose k^2=t_{k-1}+t_k.
      Then (k+1)^2=t_k+t_{k+1}=? ... how do I use the supposition?
    In my opinion, even a drawing should suffice in this case! Take a square like this

    * * * *
    * * * *
    * * * *
    * * * *

    and split it up into two triangles along the main diagonal, with one of the triangles including the diagonal.
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  7. #7
    Member courteous's Avatar
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    This is just a special case of t_3+t_4.
    Guess I learned something about proving things ... and not trying to prove the obvious.
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  8. #8
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    Quote Originally Posted by courteous View Post
    Done that, just didn't write it down. The Halmos \square sign (with a "?" mark) indicates whether this is enough ... or is induction also needed (even if it is not, still help me with the induction step ).

    MOD request: Can you please restore my spam-hijacked question?
    You have given a very nice proof...

    Suppose you were unaware of the proof you gave in your first post and were feeling lazy...

    Here is the "Proof By Induction" method

    The triangular numbers are 1, 3, 6, 10, 15, 21, 28,....

    1=1^2

    1+3=4=2^2

    3+6=9=3^2 etc

    P(k)

    t_{k-1}+t_k=k^2


    P(k+1)

    t_k+t_{k+1}=(k+1)^2

    Try to show that P(k) being valid (even if we don't know whether it is or not)
    will cause P(k+1) to be valid.

    Hence write P(k+1) in terms of P(k).


    Proof

    t_k+t_{k+1}=t_k+t_{k-1}+k+(k+1)

    =k^2+2k+1 if P(k) really is valid

    =(k+1)^2

    Now the line of dominoes is in place, hence you need to check if the first one falls.

    1+3=4=2^2 true

    0+1=1=1^2 true, depending on where you want to start from.
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