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Inductive proof foundations of analysis

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the following. 4+1+4^-1+4^-2...4^-k = 16/3-1/(3)(4^k)


    2. Relevant equations



    3. The attempt at a solution
    I confirmed the basis step and proceeded by changing the variable to n and adding n+1. I now have (16/3) - (1/(3)(4^k)) + (1/4^k+1) = 16/3 - (1/(3)(4^k+1)). In trying to get these even I have tried simplifying (that made a mess), adding and subtracting by (3)(12k), multiplying and dividing by 4^k+1 and a combination of these tactics. I just need some direction on how to proceed or to know if I am going about this completely wrong. Thank you.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 2, 2012 #2

    SammyS

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    Hello inovermyhed. Welcome to PF !

    Making your post easier to read, will likely get you more help, faster.

    Your post would be much easier to read, if you would use appropriate spacing --- even better yet, in addition to that, use the X2 & X2 icons for super/sub scripts --- even better yet use LaTeX.

    I.E.:
    [ QUOTE ]
    1. The problem statement, all variables and given/known data
    Prove the following. 4+1+4-1+4-2...4-k = 16/3-1/(3)(4k)[ /QUOTE ]​
    I suppose the (4k) should actually be in the denominator. ???
    [ QUOTE ]
    2. Relevant equations

    3. The attempt at a solution
    I confirmed the basis step and proceeded by changing the variable to n and adding n+1.

    I now have
    (16/3) - (1/(3)(4k)) + (1/4k+1) = 16/3 - (1/(3)(4k+1)). ​
    In trying to get these even I have tried simplifying (that made a mess), adding and subtracting by (3)(12k), multiplying and dividing by 4^k+1 and a combination of these tactics. I just need some direction on how to proceed or to know if I am going about this completely wrong. Thank you.
    [/QUOTE ]​

    "QUOTE" tags intentionally deactivated so OP can see what's behind the spacing & formatting.
     
  4. Sep 2, 2012 #3
    This is coming from another Real Analysis student so I may be making a mistake, but the practice for me is definitely worth it so please point out if I did something wrong.
    Taking k=1 we see that the equality holds, so we assume the equality holds for k and attempt to prove for k+1. We then have [itex] {\frac{16}{3}} - {\frac{1}{3 * 4^k}} + 4^{-k -1} = {\frac{16}{3}} - {\frac{1}{3 * 4^{k+1}}} [/itex]
    Subtract 16/3 from both sides and multiply by negative 3 to get
    [itex]{\frac{1}{{4^k}}} - 3 * 4^{-k -1} = {\frac{1}{4^{k+1}}} [/itex]
    You can then multiply both sides by [itex] 4^{k+1} [/itex] to get
    [itex] {\frac{4^{k+1}}{4^k}} - 3*4^{k+1} * 4^{-k -1} = 1 [/itex]
    but the first term on the left just gives 4 and through a little manipulation i.e breaking [itex] 4^{k+1} [/itex] into [itex] 4^k *4 [/itex] and a similar maneuver with [itex] 4^{-k -1}[/itex] you get 4-3 = 1 thus proving the equality.
    I think that is correct.
     
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