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Math Help - Mathematical Induction Question

  1. #1
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    Mathematical Induction Question

    Hello,
    These problems are so easy but I am making it too hard... I can get everything up to the proof part, which is underlined below

    his is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.

    I know this is "factoring out the k+1" but I dont understand what that means.... could someone explain or break it down? Thanks!!
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  2. #2
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    Quote Originally Posted by triathlete View Post
    Hello,
    These problems are so easy but I am making it too hard... I can get everything up to the proof part, which is underlined below

    his is not too hard: 1 + 2 + ... + k + (k+1) = k(k+1)/2 + (k+1) = (k(k+1) + 2 (k+1))/2 = (k+1)(k+2)/2. The first equality is a consequence of the inductive assumption.

    I know this is "factoring out the k+1" but I dont understand what that means.... could someone explain or break it down? Thanks!!
    You would agree that xy+xz=x(y+z)? Let x=k+1,y=k,z=2
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  3. #3
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    That makes sense.... what about this

    (k(k+1)(2k+1) + 6(k+1)^2)/6 = ((k+1)[k(2k+1) + 6(k+1)])/6

    In my book, it has this step and it says "by factoring out (k+1)". I guess I just don't see it here, how does the "k" move to before the 2k+1 and how does the square get removed??
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  4. #4
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    Quote Originally Posted by triathlete View Post
    That makes sense.... what about this

    (k(k+1)(2k+1) + 6(k+1)^2)/6 = ((k+1)[k(2k+1) + 6(k+1)])/6

    In my book, it has this step and it says "by factoring out (k+1)". I guess I just don't see it here, how does the "k" move to before the 2k+1 and how does the square get removed??
    You cannot do it by factoring out in this particular case.

    We are speaking of the principle of mathematical induction, which is the consequence of the Well-Ordered Principle.

    Theorem says:

    For each positive integer n, let P(n) be a statement. If

    (1) p(1) is true and
    (2) the implication P(k) \Rightarrow P(k+1) is true for every positive integer k,

    then P(n) is true for every positive integer n.

    This is an analogy of climbing a ladder to whatever height you want.

    If you a have lover waiting for you by the window in an breachable castle, say you are Romeo, and your lover is Juliet. You need a good ladder. If you cannot get to the first rung, you cannot get you Juliet. Say, if you could to get to the first rung but the subsequent rungs are missing, you still cannot get to embrace Juliet. So you must have a good ladder with every rung being intact provided you can climb.
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  5. #5
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    Quote Originally Posted by triathlete View Post
    That makes sense.... what about this

    (k(k+1)(2k+1) + 6(k+1)^2)/6 = ((k+1)[k(2k+1) + 6(k+1)])/6

    In my book, it has this step and it says "by factoring out (k+1)". I guess I just don't see it here, how does the "k" move to before the 2k+1 and how does the square get removed??
    Look at the two parts to that sum:
    the first term is
    k(k+1)(2k+1)

    If we "take out" the k+ 1 it leaves k(2k+1)

    the second term is 6(k+1)^2= 6(k+1)(k+1)

    If we take out" one of the two k+ 1 terms, it leaves 6(k+1).

    That is k(k+1)(2k+ 1)+ 6(k+1)^2= (k+1)(k(2k+1)+ 6(k+1).
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