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Math Help - Induction

  1. #1
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    Induction

    I want to prove \sum_{k=1}^{n}k=\frac{n(n+1)}{2}=<br />
\sum_{k=0}^{n}(-1)^{n+k}*k^2

    So, simply what I chose to do is to first prove
    \sum_{k=1}^{n}k=\frac{n(n+1)}{2}
    by (mathematical) induction, and then use the same method for
    \frac{n(n+1)}{2}=\sum_{k=0}^{n}(-1)^{n+k}*k^2
    The first works out fine, no problem there at all.
    The second part causes problems.

    I manage to show P(x) true for x=1 (wow, I am realy amazing)
    And then I make my claim:
    \sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2
    I then do my calculations and I get:

    \frac{3m_0+5m_0+2}{2}
    Witch is not right.
    Anyone know or have a clue about where I am off?
    Is there another part of my calculations that I should include? Please say so if that is the case, I try to include just the parts where I think I am wrong (everything only if I have no clue).
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by a4swe
    I want to prove \sum_{k=1}^{n}k=\frac{n(n+1)}{2}=<br />
\sum_{k=0}^{n}(-1)^{n+k}*k^2

    So, simply what I chose to do is to first prove
    \sum_{k=1}^{n}k=\frac{n(n+1)}{2}
    by (mathematical) induction, and then use the same method for
    \frac{n(n+1)}{2}=\sum_{k=0}^{n}(-1)^{n+k}*k^2
    The first works out fine, no problem there at all.
    The second part causes problems.

    I manage to show P(x) true for x=1 (wow, I am realy amazing)
    And then I make my claim:
    \sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2
    I then do my calculations and I get:

    \frac{3m_0+5m_0+2}{2}
    Witch is not right.
    Anyone know or have a clue about where I am off?
    Is there another part of my calculations that I should include? Please say so if that is the case, I try to include just the parts where I think I am wrong (everything only if I have no clue).
    Try:

    Assume for some integer m_0 that:

    <br />
\frac{m_0(m_0+1)}{2}=\sum_{k=0}^{m_0}(-1)^{m_0+k}*k^2<br />

    Then:

    <br />
\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2 <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1+k}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />
    <br />
=-\sum_{k=0}^{m_0}(-1)^{m_0+k}*k^2+(m_0+1)^2<br />
    .

    Then by assumption the first term on the right may be replaced by -\frac{m_0(m_0+1)}{2} so:

    <br />
\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k}*k^2 <br />
=-\frac{m_0(m_0+1)}{2}+(m_0+1)^2<br />
.

    Which should simplify to what you are looking for to prove the required equality for m_0+1.

    RonL
    Last edited by CaptainBlack; September 4th 2006 at 02:44 PM.
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  3. #3
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    Well then, there is something that I don not understand here:

    why:
    <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1+k}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />

    and not:

    <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by a4swe
    Well then, there is something that I don not understand here:

    why:
    <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1+k}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />

    and not:

    <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />
    Because the first of these is the far right term in the original problem
    with n replaced by m_0+1, that is the next case after the one
    assumed to hold.

    RonL
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  5. #5
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    Oh, sorry a little typo again I mean to say:

    Why:

    <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1+k}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />

    And not:

    <br />
=\sum_{k=0}^{m_0+k}(-1)^{m_0}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />

    'cuse I did take the m_0+1 term for it self and only got m_0 left...right?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by a4swe
    Oh, sorry a little typo again I mean to say:

    Why:

    <br />
=\sum_{k=0}^{m_0}(-1)^{m_0+1+k}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />

    And not:

    <br />
=\sum_{k=0}^{m_0+k}(-1)^{m_0}*k^2 + (-1)^{2(m_0+1)}(m_0+1)^2<br />

    'cuse I did take the m_0+1 term for it self and only got m_0 left...right?
    It's the same answer as before. The first of these is:

    <br />
\sum_{k=0}^{m_0}(-1)^{m_0+k} k^2<br />

    with m_0 replaced by m_0+1 throughout, and
    the last term taken outside the sum:

    <br />
\sum_{k=0}^{m_0+1}(-1)^{m_0+1+k} k^2=\left[ \sum_{k=0}^{m_0}(-1)^{m_0+1+k} k^2 \right] +  (-1)^{m_0+1+m_0+1} (m_0+1)^2<br />


    RonL
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  7. #7
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    Well, yes off course.
    Hmmm, but why do one use n+k?
    Isn't there any way to express the same thing with fix numbers, for me it seems like that should be possible, but probably I am wrong.
    Thanks a lot CB, you are often of great help.
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