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Induction Proof Help...

  1. Feb 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG
    2. Relevant equations
    Prof. Note's. Capturev 1.PNG
    3. The attempt at a solution
    I'm on the 3 line where my Prof. combines both equations, I'm confused on what my equation should look. Her's was (n+1)(n+1)+1)/2
    20160206_163947.jpg
     
  2. jcsd
  3. Feb 6, 2016 #2

    HallsofIvy

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    Saying that "4 divides [itex]5^n- 1[/itex] is NOT just a "reference" to [itex]\frac{5^n- 1}{4}[/itex]! It is the statement that [itex]\frac{5^n- 1}{4}[/itex] is an integer. That is, [itex]\frac{5^n- 1}{4}= k[/itex] for some integer k. Further the "n+1" form of the formula is not [itex]\frac{5^{n+1}- 1}{4}+ (n+1)[/itex]. I don't where you got that additional "(n+1)"! Replacing n by n+1 in [itex]\frac{5^{n}- 1}{4}[/itex] is just [itex]\frac{5^{n+1}- 1}{4}[/itex].

    Now, of course, you want to "algebraically" go back to the "[itex]5^n[/itex]" and to do that use the fact that [itex]5^{n+1}= 5(5^n)[/itex].
    It will be helpful to use [itex]5(5^n)- 1= 5(5^n)- 5+ 4[/itex].
     
  4. Feb 6, 2016 #3

    ehild

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    You can write 5n+1 as 5 * 5n, and 5n+1 - 1 = 5 * 5n -5 +4 = 5(5n-1) + 4.
     
  5. Feb 6, 2016 #4
    Sorry, I just noticed this but should my equation be 4 l 5^n-1 or is what have okay?
     
  6. Feb 6, 2016 #5

    ehild

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    Yes. As 5n-1 is divisible by 4 , the first therm of 5(5n-1) + 4 is divisible by 4, and the second term is just 4.
     
  7. Feb 6, 2016 #6

    HallsofIvy

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    "4 l 5^n-1" is NOT even an equation!
     
  8. Feb 6, 2016 #7
    Okay sorry, I'm a bit confused now. So (5^(n)-1)/4 is correct, right ? And would I follow what my prof. did, so I set my work for n+1 to 5(5n-1) + 4 or is that the induction step?
     
  9. Feb 6, 2016 #8
    Sorry, I just refreshed my page and your comment appeared, thank you for the help :)
     
  10. Feb 15, 2016 #9
    I think that you have to go from the fact that if (5^n)-1 is divisible by 4 then (5^n)-1=4k where k is a constant. Now, how can you apply this to 5((5^n)-1)+4?. Think of substitution
     
    Last edited: Feb 15, 2016
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