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Simple series confusion

  1. Jun 6, 2012 #1
    s5enh3.png

    I understand that they have done [tex]\displaystyle\sum_{r=k-1}^{2k} r =\sum_{r=1}^{2k}r - \sum_{r=1}^{k-1-1}r[/tex]

    But where did r = 1 come from as r = k-1 before is what I don't understand.

    Also this question:

    Show that [tex] \displaystyle\sum_{p=3}^{n}(4p+5) = (2n+11)(n-2) [/tex]
    [tex] \displaystyle4\sum_{p=1}^{n}p + 5\sum_{p=1}^{n}1 = ... [/tex]

    How do you go from p=3 to p=1
     
  2. jcsd
  3. Jun 6, 2012 #2
    I'll break this down to simple numbers, but the same logic applies for your problem.

    Lets say you are asked for the sum [itex]3+4+5[/itex]. Doing this in summation form, you could add the numbers from 1 to 2 i.e [itex]1+2[/itex] and subtract it from the sum of 1 to 5, i.e [itex]1+2+3+4+5[/itex], which gives you the same result.

    Other way of seeing this is, say you are asked to find the sum [itex]S[/itex]. You introduce another quantity [itex]Q[/itex] such that you know the value of [itex]S+Q[/itex] and [itex]Q[/itex]. Subtracting these known values, you can find [itex]S[/itex].
     
  4. Jun 6, 2012 #3
    I can see how it applies to the first question, but not second going from p=3 to p=1 makes no sense to me.
     
  5. Jun 6, 2012 #4
    Well, the second question is wrong. Or you might not have read it properly.

    That's why I didn't include it in my quote :tongue:
     
  6. Jun 6, 2012 #5
    @ Infinitum, wait, really? Because Wolfram Alpha and I both got the provided answer...

    @ synkk: [itex]\displaystyle \sum_{p=1}^{n} 4p + 5 = \displaystyle \sum_{p=1}^{2} 4p + 5 + \displaystyle \sum_{p=3}^{n} 4p+5[/itex]
     
  7. Jun 6, 2012 #6
    got it guys thanks
     
  8. Jun 7, 2012 #7
    The summation you are trying to prove is of course correct. What's wrong is....

    [tex] \displaystyle\sum_{p=3}^{n}(4p+5) = \displaystyle4\sum_{p=1}^{n}p + 5\sum_{p=1}^{n}1 [/tex]
     
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