Question

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?

Answer 1

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

Answer 2

So you have never dealt with double-sigma notation before?

Answer 3

i have but not when the inner index depends on the outer one
I had examples like \[\sum_{i=1}^n\sum_{j=1}^m\]

Answer 4

FYI, I tried n=3 on Wolfram and it gives 20 as an answer, but still doesn't help me.

Answer 5

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

Answer 6

Do you have to find the sum for general \(n\), or just any \(n\)?

Answer 7

Oh I'm just asking for any n. I just need an example of how to compute the double sum.

Answer 8

In particular, I was wondering maybe how would we calculate it for say n=3

Answer 9

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.

Answer 10

To be able to always compute the double sums correctly, the upper limit for the second sigma must be \(\bf \ge n + 1\)

Answer 11

Oh I see o.o Is there a systematic way of doing this for large n?

Answer 12

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.