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Homework Help: Summation Question

  1. Sep 14, 2010 #1
    heMwm.png

    So whats going on here? Since there is a 2^(k-1), I can subtract one from n and also the index? Thats what it looks like they did. Also, where did they get that 1/2 from?
     
  2. jcsd
  3. Sep 14, 2010 #2

    danago

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    The 1/2=2-1 is simply the term of the summation for which k=-1; by bringing it out of the summation, the lower bound can be changed from k=-1 to k=0
     
  4. Sep 14, 2010 #3
    Oh ok..so how is it that we can change the bound? What is the rule for that or steps?
     
  5. Sep 14, 2010 #4

    danago

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    Think about it like this:

    [tex]\sum\limits_{k = 0}^{n + 1} {{2^{k - 1}}} = {2^{ - 1}} + {2^0} + {2^1} + ... + {2^{n - 1}} + {2^n} = \sum\limits_{k = - 1}^n {{2^k}} [/tex]

    Do you see why changing the bounds is valid?
     
  6. Sep 14, 2010 #5
    But the bound for the first was n+1, so shouldnt the last term be 2^(n+1) or no?
     
  7. Sep 14, 2010 #6

    danago

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    But remember, the index of 2 is now k-1. So when k=n+1, the index will be (n+1)-1=n
     
  8. Sep 14, 2010 #7
    So if my lower bound were 1, I could bring out a 2 and the new index would start from 2? And if the bound were 2, I could bring out a 4 and new bound would be 3?
     
  9. Sep 14, 2010 #8

    danago

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    Yes thats right
     
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