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Simple geometric series question

  1. Jun 16, 2014 #1
    Take the case for the mean:

    \bar{x} = \frac{1}{N} \Big( \sum_{i=1}^Ni \Big)

    If we simply use the formula for the sum of a geometric series, we get

    [itex]\bar{x} = \frac{N}{2} (2a + (N - 1)d)[/itex]

    where a and d both equal 1, so we simply get the result

    [itex]\bar{x} = \frac{1}{2} (N + 1)[/itex]

    What I've been trying to do is to get the same result by being more rigorous, in that I've attempted to expand the series, i.e.

    \bar{x} = \frac{1}{N} (1 + 2 + 3 + ... + N)

    and replace the terms with the appropriate expressions using N, for example 1 and 2 would be N-(N-1) and N-(N-2) respectively, and so forth. Unfortunately I just keep going round in circles and never achieve the correct result.

    Am I wasting my time? Or have I simply not seen the next step?
  2. jcsd
  3. Jun 16, 2014 #2
    The idea is to not replace all the terms, but only have the terms. For example ##2## gets replaced by ##N - (N - 2)##, but we don't want to replace the later term ##N-2##. The terms will cancel out then.
  4. Jun 16, 2014 #3

    I think I understand, but just to be sure, could you clarify further please?
  5. Jun 16, 2014 #4
    Clarify what? What is unclear about my explanation? Just take your sum ##1+2+3+...+N## and replace the first half of the terms in the way you indicated and leave the other half.
  6. Jun 16, 2014 #5
    I'm still eagerly awaiting the question about a geometric series! :tongue2:
  7. Jun 16, 2014 #6
    Well I've already attempted what you've suggested, so it seems, though I cannot reach the final answer, as I've stated already.

    I've tried going through the puzzle again and I sometimes obtain N/2 + 1, or N/2 as my final answer instead of (N+1)/2. I'm really confused as to where the +1 comes from. Somehow the final answer evades me.
  8. Jun 16, 2014 #7
    The idea is to do this:

    [tex]1+2+3+4+5 = (5-4) + (5-3) + 3 + 4 + 5 = (5-4) + 4 + (5-3) + 3 + 5 = 5 + 5 + 5= 15[/tex]
  9. Jun 17, 2014 #8
    This looks like an arithmetic sequence to me.
  10. Jun 17, 2014 #9
    Well, as is usually the case, I left the problem yesterday evening and decided to sleep on it. This morning I woke up and solved it within a minute. Don't you just love it when you hit a wall and just jump over it the next time you look at a problem from a different angle?

    Thanks to all who helped.
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