kirbykirby
  • kirbykirby
Double sum question? An example to calculate \[\sum_{i=1}^n\sum_{j=i+1}^n(i+2j)\]?? I only have an example where n=1 and it gives a sum of 0 (why?) Maybe with n=3, what would the expanded form look like?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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kirbykirby
  • kirbykirby
I think I'm a bit confused because of the "i" also present in the inner sum's index :S I haven't encountered this yet
anonymous
  • anonymous
So you have never dealt with double-sigma notation before?
kirbykirby
  • kirbykirby
i have but not when the inner index depends on the outer one I had examples like \[\sum_{i=1}^n\sum_{j=1}^m\]

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kirbykirby
  • kirbykirby
FYI, I tried n=3 on Wolfram and it gives 20 as an answer, but still doesn't help me.
kirbykirby
  • kirbykirby
http://www.wolframalpha.com/input/?i=Sum%5Bi%2B2j%2C+%7Bi%2C+1%2C+3%7D%2C+%7Bj%2C+i%2B1%2C+3%7D%5D
anonymous
  • anonymous
Do you have to find the sum for general \(n\), or just any \(n\)?
kirbykirby
  • kirbykirby
Oh I'm just asking for any n. I just need an example of how to compute the double sum.
kirbykirby
  • kirbykirby
In particular, I was wondering maybe how would we calculate it for say n=3
anonymous
  • anonymous
The thing is that for lower values of n, you could create an array/table of i and j. Although when n = 3, and j = i + 1 and i = 3, then j = 4 but it only goes up to n which is 3. And this is where the conflict comes in.
anonymous
  • anonymous
To be able to always compute the double sums correctly, the upper limit for the second sigma must be \(\bf \ge n + 1\)
kirbykirby
  • kirbykirby
Oh I see o.o Is there a systematic way of doing this for large n?
kirbykirby
  • kirbykirby
Ok So I have slept and gone back over this question. I realized that adding the j term is like adding the sum of consecutive integers formula, but not starting at an index of 1. \[\sum_{i=1}^n \left( \sum_{j=i+1}^{n}i +2\sum_{j=i+1}^{n}j \right)\] \[\sum_{i=1}^n \left(i(n-(i+1)+1) +2\sum_{j=1}^{n-i}(j+i) \right)\] by index shifting \[\sum_{i=1}^n \left(i(n-i) +2\sum_{j=1}^{n-i}j+2\sum_{j=1}^{n-i}i \right)\] \[\sum_{i=1}^n \left(i(n-i) +2*\frac{(n-i)(n-i+1)}{2}+2*(i(n-i-1+1)) \right)\]using the sum of consectuive integers formula on j \[\sum_{i=1}^n \left(i(n-i) +(n-i)(n-i+1)+2i(n-i) \right)\] \[\sum_{i=1}^n \left((n-i)(i+n-i+1+2i) \right)\] \[\sum_{i=1}^n \left((n-i)(n+1+2i) \right)\] Am I on the right track? If so the rest is easy.

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