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Evaluating the sum of a sigma notation problem with a lower limit k=10

  1. Dec 9, 2012 #1
    How do I evaluate the sum of this sigma notation problem?

    20
    ∑ k
    k=10

    Normally, I would think to use the theorem for the sum of the first n integers:
    n
    ∑ k = n(n+1)/2
    k=1

    I don't know how to do this, however, since the lower limit is k=10, not k=1.

    My professor wrote this note on the board for the problem. It gives the answer, but I still don't understand it. If you do, could you explain? Thanks!

    20....20....9
    ∑ k = ∑ - ∑ k
    k=10 k=1 k=1

    20(21)/2 - 9(10)/2 = 165 <-- (The answer)
     
  2. jcsd
  3. Dec 9, 2012 #2
    I think I've got an answer from another responder. His post is below:

    For this sum you can just plug in the values until you hit 20. That's all you need to do.

    10+11+12+13+14+15+16+17+18+19+20 = 165

    His way is unnecessary and requires much more work. A better way is this, when ever you subtract k from the lower limit and the upper limit, you need to add that k in the sigma notation. In this way you can put your lower limit back to 1.

    so

    20....19 .......... 11
    ∑ k = ∑ (k+1) = ∑ (k+9) = 165
    k=10 k=9 .......... k=1
     
  4. Dec 9, 2012 #3

    haruspex

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    That's pretty straightforward. It just says that the sum of the first 20 terms of a series can be split into the sum of the first 9 and the sum of the next 11.
     
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