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Induction for sum = 2^(n+2)+2

  1. Jul 7, 2014 #1
    1. The problem statement, all variables and given/known data

    prove by induction [tex]\sum_{j=1}^{n+1} j \cdot 2^j = n \cdot 2^{n+2}+2; n \ge 0[/tex]





    2. The attempt at a solution

    P(0)

    [tex]\sum_{j=1}^{0+1} j \cdot 2^j = 0 \cdot 2^{0+2}+2[/tex]

    [tex] 2+2 [/tex]

    here is where I need some help

    is P(k)

    [tex]\sum_{j=1}^{k+1} j \cdot 2^j = (k+1) \cdot 2^{k+3}+2[/tex] ??

    then

    P(k+1) [tex]\sum_{j=1}^{k+2} j \cdot 2^j = (k+2)\cdot 2^{k+4}+2[/tex]
     
    Last edited: Jul 7, 2014
  2. jcsd
  3. Jul 7, 2014 #2

    LCKurtz

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    No. Copy your original problem (the first equation above) exactly with ##n = k##.
     
  4. Jul 8, 2014 #3
    [tex]\sum_{j=1}^{k} j \cdot 2^j = (k) \cdot 2^{k+2}+2[/tex] why this when the sum goes to n+1??
     
  5. Jul 8, 2014 #4

    LCKurtz

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    What part of "copy it exactly" don't you understand?
     
  6. Jul 8, 2014 #5
    I see what I did wrong in my reply. it was a simple mistake, hostility doesn't help. I'll figure it out. thanks
     
  7. Jul 8, 2014 #6

    Fredrik

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    You want to prove that for all non-negative integers n, we have ##\sum_{j=1}^{n+1} j \cdot 2^j = n \cdot 2^{n+2}+2##. For each non-negative integer n, let P(n) be the statement ##\sum_{j=1}^{n+1} j \cdot 2^j = n \cdot 2^{n+2}+2##. Let the scope of the ##\forall## symbol ("for all") be the set of non-negative integers. The result that you want to prove can now be stated as
    $$\forall n~P(n)$$ To prove this is to prove all of the infinitely many statements P(0), P(1), P(2),... The idea behind induction proofs is that you can do that in a finite number of steps. All you have to do is to prove the following two statements:
    \begin{align}
    &P(0)\\
    &\forall k~\big(P(k)\Rightarrow P(k+1)\big)
    \end{align} When you get to the second step, you don't really have to think about what P(k) or P(k+1) is. You just use the definition. It's appropriate to think of P as a kind of "function" that takes integers to statements.

    I would recommend that you always write down a definition of P(k) for all the relevant k when you start an induction proof. You know that your definition is OK if it turns the statement that you want to prove into ##\forall n~P(n)##. The theorem has to be a statement of this form, if induction is to be used.
     
    Last edited: Jul 8, 2014
  8. Jul 8, 2014 #7

    LCKurtz

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    It isn't hostility. It is my expressing frustration with with your carelessness in executing an extremely simple task.
     
  9. Jul 8, 2014 #8
    It wasn't carelessness. Ive been working on my schoolwork for 16 hours a days for the past month. It was a simple mistake. I appreciate the help.
     
  10. Jul 8, 2014 #9

    LCKurtz

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    So, after you wrote down P(k) and P(k+1), were you able to show P(k) implies P(k+1)?
     
  11. Jul 8, 2014 #10
    so p(k) is

    [tex]\sum_{j=1}^{k+1} = k \cdot 2^{k+2} +2[/tex]

    so p(k+1)

    is

    [tex] \sum_{j=1}^{k+2} = (k+1) \cdot 2^{k+3}+2 [/tex]

    shifting the index

    [tex]\displaystyle{\sum_{j=1}^{k+2}}j 2^j=\displaystyle{\sum_{j=1}^{k+1}}j 2^j+(k+2) \cdot 2^{k+2}[/tex]

    correct?

    now

    [tex]\displaystyle{\sum_{j=1}^{k+1}}j 2^j+(k+2)2^{k+2}=(k \cdot 2^{k+2}+2) + (k+2)2^{k+2}[/tex]

    [tex] = k \cdot 2^{k+2}+k \cdot 2^{k+2} + 2 \cdot 2^{k+2} + 2 [/tex]

    [tex]=2\cdot k \cdot 2^{k+2} + 2 \cdot 2^{k+2} + 2 [/tex]

    [tex] = (2k+2)2^{k+2}+2[/tex]

    [tex] = (k+1)2^{k+3} + 2[/tex]

    ??
     
  12. Jul 8, 2014 #11

    LCKurtz

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    I don't think you need the ??'s at the end. I think you know it is correct.
     
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