1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simple series confusion

  1. Jun 6, 2012 #1

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook