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Math Help - Bacteria by Induction

  1. #1
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    Bacteria by Induction

    I am trying to prove this by induction, but have so far failed.

    I note that:

    b1 = 1, r1 = 7
    b2 = 7, r2 = 13
    b3 = 13, r3 = 55

    But from there i am stuck. As always thanks to all.

    A new species of bacteria called blue and red behave in the following way: after one minute every blue bacterium turns into 6 red ones, and every red bacterium turns into 1 blue and 1 red bacterium. During an experiment you put 1 blue and 1 red bacterium into a test-tube, so that after one minute there were 1 blue bacterium and 7 red ones; and after two minutes there were 7 blue ones and 13 red ones. Prove that after n minutes there will be:

    3/5 * 3^n + 2/5 * (-2)^n
    Blue bacteria in the test-tube.
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  2. #2
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    Here is one way.

    Prove by induction:
    bn = (3/5)[3^n] +(2/5)[(-2)^n]

    Okay.
    Let us see first if b1 is true.
    Since you found b1 = 1, we see if that is true.
    When n = 1,
    b1 = (3/5)[3^1] +(2/5)[(-2)^1]
    b1 = (3/5)[3] +(2/5)[-2]
    b1 = 9/5 -4/5
    b1 = 5/5 = 1 ----it is true.

    Then let us assume when n = k,
    bk = (3/5)[3^k] +(2/5)[(-2)^k] is true.

    Then if we can show that when n = k+1,
    b_(k+1) = (3/5)[3^(k+1)] +(2/5)[(-2)^(k+1)] is true also,

    then we can say that
    bn = (3/5)[3^n] +(2/5)[(-2)^n] is really true.
    It is proven by mathematical induction.

    So,
    b_(k+1) = (3/5)[3^(k+1)] +(2/5)[(-2)^(k+1)]
    b_(k+1) = (3/5)[3^k *3^1] +(2/5)[(-2)^k *(-2)^1]
    b_(k+1) = (3/5)[3^k *3] +(2/5)[(-2)^k *(-2)]
    b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)

    Now we use your findings.
    You found
    b2 = 7
    b_(2+1) = b3 = 13

    Plugging those into the Eq.(i), when k=2,
    b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)
    b_(2+1) = (9/5)[3^2] -(4/5)[(-2)^2]
    b_(2+1) = (9/5)[9] -(4/5)[4]
    b_(2+1) = 81/5 -16/5
    b_(2+1) = 65/5 = 13
    Or,
    b3 = 13 ---which is true.

    Therefore, bn = (3/5)[3^n] +(2/5)[(-2)^n] is proven.
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  3. #3
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    Very much appreciated ticbol!

    However there is one thing which i want to clear up, in the end when you sub in b2 = 7, can you do that to prove it for all cases, i was wondering if that only proved it for the 2nd case, and not the kth + 1 case.

    Any clarification would be great.
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  4. #4
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    I tried not using your findings but I could not find a way.
    After
    b_(k+1) = (9/5)[3^k] -(4/5)[(-2)^k] ----(i)
    I was stuck.
    If bn were only a summation or sum, then maybe I could have found a way.
    But since bn is defined differently, I was forced to use two of your findings.

    By the way, I found also your findings by brute force, by diagrams.

    Tomorrow, when I have time, I will try again.

    ----------
    Umm, I think I missed your point.
    So, you were asking if my using k=2 will cover all k+1?

    Sure it will.
    Remember, if k=2, then k+1 = 2+1 = 3
    So if b3 is true by using k=2, then k and k+1 are satisfied.
    We only need to satisfy one k+1 for the induction.

    If you like, test this case.
    k=1, and so k+1 = 2.
    You will get b2 = 7 by using k=1.

    Or,
    k=3, anmd so k+1 = 4.
    You will find b4 = 55 by using k=3.
    Last edited by ticbol; June 20th 2005 at 05:42 AM.
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  5. #5
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    Thanks a lot ticbol. Can I inquire where you are located?

    Thanks also for spending so much time on this question.
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  6. #6
    Site Founder Math Help's Avatar
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    Thank you ticbol

    Ticbol thank you for providing such detailed answers you are a real asset to all of the members!

    Please let me invite you both to our introductions thread if you care to share with us any information about yourselves.
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  7. #7
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    Umm.

    Well, I am a Filipino civil engineer working in constructions here in Guam.

    I love Math. Like chess, it is one way for me to unwind.
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  8. #8
    Site Founder MathGuru's Avatar
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    The big knight switch

    Like chess huh?

    Have you seen The Big Knight Switch ?

    I have spent about 20 minutes on it and I haven't solved it yet.
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  9. #9
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    Sorry, MathGuru, but that is not chess.
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