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Basic Mathematical Induction w/ n In the Exponent

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    All I am trying to do is the basic mathematical induction routine: n=1, n=k, n=k+1 and how n=k proves n=k+1. The problem I am having is with the algebra.


    2. Relevant equations and Attempts

    For n=1

    [tex]4=4[/tex] That's OK

    For n=k


    For n=k+1 by Substitution


    To Prove by Adding [tex]K+1[/tex] to [tex]S_{k}[/tex]


    It took me a minute to figure out how [tex]\frac{4}{3}(4^{k}-1)[/tex] became [tex]\frac{4^{(k+1)}}{3}[/tex], but I got that part.

    [tex]= \frac{4^{(k+1)}}{3}+4^{(k+1)}-\frac{4}{3}[/tex]

    This is where I am stuck. I gather that I am supposed to multiply [tex]4^{(k+1)}[/tex] by [tex]\frac{3}{3}[/tex], and then combine like terms. I am confused about this because I don't know how to multiply the term by 3 (I probably learned it, I'm just stumped).

    Please help me understand what to do next. As long as I can understand how to manipulate the exponential terms with the fractions, I can complete this. Thanks for your time.
  2. jcsd
  3. Dec 9, 2008 #2
    Check this expression.
    Once you have the correct target expression, the next part should work out.
  4. Dec 9, 2008 #3


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    ummm....when you multiply [tex]4^{k+1}[/tex] by [tex]\frac{3}{3}[/tex] you just get [tex]\frac{3(4^{k+1})}{3}[/tex]....and since you now have two terms with the same denominator, you can add the numerators to get: [tex]\frac{4^{(k+1)}+3(4^{(k+1)})}{3}-\frac{4}{3}[/tex]

    what do you get when you add one of 'something' to 3 of 'something'?
  5. Dec 9, 2008 #4
    I'm not sure how that is incorrect.

    I get it now. When I look at it like 1[4^(K+1)]+3[4^(K+1)], it makes sense. I was trying to multiply, which was wrong, and didn't let me pull out 4/3.

    Thanks, everyone.
  6. Dec 9, 2008 #5
    When you substitute n=k+1,
    right hand side of the expression should be [tex]4/3[/tex]([tex]4^{k+1}[/tex] -1)
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