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Math Help - Mathematical Induction Arithmetic

  1. #1
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    Mathematical Induction Arithmetic

    I was working on proving using mathematical induction the premise:

    P(n): 2 - 2*7 + 2*7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4 while n>=0

    I got to the point of proving P(n+1),

    [1-(-7)^(n+1) + 8(-7)^(n+1)]/4

    but I'm not sure arithmetically how that will simplify into the expected outcome of:

    [1-(-7)^(n+2)]/4

    ???

    Am I even right that it will simplify to that seeing as how that is what I was expecting out of the original P(n).
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  2. #2
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    Quote Originally Posted by buddyp450 View Post
    I was working on proving using mathematical induction the premise:

    P(n): 2 - 2*7 + 2*7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4 while n>=0

    I got to the point of proving P(n+1),

    [1-(-7)^(n+1) + 8(-7)^(n+1)]/4

    but I'm not sure arithmetically how that will simplify into the expected outcome of:

    [1-(-7)^(n+2)]/4

    ???

    Am I even right that it will simplify to that seeing as how that is what I was expecting out of the original P(n).
    2-2(7)+2(7^2)+..........+2(-7)^n=\frac{1-(-7)^{n+1}}{4} ??

    If so, then

    2-2(7)+........+2(-7)^n+2(-7)^{n+1}\ must\ =\frac{1-(-7)^{n+2}}{4}

    Check:

    \frac{1-(-7)^{n+2}}{4}=\frac{1-(-7)^{n+1}(-7)}{4}=\frac{1-(-7)^{n+1}}{4}+\frac{8(-7)^{n+1}}{4} =\frac{1-(-7)^{n+1}}{4}+2(-7)^{n+1}

    Hence, if true for some n, then definately true for n+1 because of that,
    so the dominoes are all stacked up.

    True for n=0 ?

    2=\frac{1-(-7)^1}{4} ?

    Yes...... Proven
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  3. #3
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    Quote Originally Posted by buddyp450 View Post
    I was working on proving using mathematical induction the premise:

    P(n): 2 - 2*7 + 2*7^2 - ... + 2(-7)^n = (1-(-7)^(n+1))/4 while n>=0

    I got to the point of proving P(n+1),

    [1-(-7)^(n+1) + 8(-7)^(n+1)]/4 <--this is correct

    but I'm not sure arithmetically how that will simplify into the expected outcome of:

    [1-(-7)^(n+2)]/4

    ???

    Am I even right that it will simplify to that seeing as how that is what I was expecting out of the original P(n).
    Working with the expression  <br />
\frac{1}{4}\{1-(-7)^{n+1}+8(-7)^{n+1)}\}<br />
, the tricky part here is the application of Distributive Law to a negative term. Now the positive integer 8 becomes negative.

     <br />
\frac{1}{4}[1-(-7)^{n+1}(1-8)]<br />
, which is  <br />
\frac{1-(-7)^{n+2}}{4}<br />
,
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