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

  • Thread starter Brianna V
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  • #1
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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)
 

Answers and Replies

  • #2
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0
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
 
  • #3
haruspex
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20....20....9
∑ k = ∑ - ∑ k
k=10 k=1 k=1
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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