1. ## Double summation

Hi

I'm having some difficulty with the following problem. Evaluate

$\displaystyle \Sigma^5_{i=1}\Sigma^4_{j=i}(2i+j+2)$

The book I'm reading doesn't have an example where the two numbers above the sigmas are different, but does have an example where they are the same. So I'm not sure how to approach this one. I had a bash but came up with the incorrect answer (book says it should be 210)

As I looked at it further I became even more confused, since i ranges from 1 to 5, then when i=5 the inner sum ranges from 5 to 4? I don't see how that makes sense

If anyone could shed any light on this or point me in the right direction to solve it, it would be much appreciated.

2. Split it up:

$\displaystyle \sum_{i=1}^5 \left({ \left({2i + 1 + 2}\right) + \left({2i + 2 + 2}\right) + \left({2i + 3 + 2}\right) + \left({2i + 4 + 2}\right)}\right)$

It's straightforward once you understand what you're doing.

3. Originally Posted by Matt Westwood
Split it up:

$\displaystyle \sum_{i=1}^5 \left({ \left({2i + 1 + 2}\right) + \left({2i + 2 + 2}\right) + \left({2i + 3 + 2}\right) + \left({2i + 4 + 2}\right)}\right)$

It's straightforward once you understand what you're doing.
Ok, to me that looks like you've taken j to range from 1 to 4, when the question says it ranges from i to 4. I think I'm even more confused now

4. Whoops, so it does ... apologies, ignore me. I'm too tired to be on this site at the moment!

Okay, so treat it as the one where i and j range over the same interval, but note that when i=5 there is no j that fits the condition. So when i=5 the summation is vacuous, and you can treat it as the same as when i goes from 1 to 4.

5. Originally Posted by Stonehambey
Hi

I'm having some difficulty with the following problem. Evaluate

$\displaystyle \Sigma^5_{i=1}\Sigma^4_{j=i}(2i+j+2)$

The book I'm reading doesn't have an example where the two numbers above the sigmas are different, but does have an example where they are the same. So I'm not sure how to approach this one. I had a bash but came up with the incorrect answer (book says it should be 210)

As I looked at it further I became even more confused, since i ranges from 1 to 5, then when i=5 the inner sum ranges from 5 to 4? I don't see how that makes sense

If anyone could shed any light on this or point me in the right direction to solve it, it would be much appreciated.
When i=1,
$\displaystyle \Sigma^4_{j=i}(2i+j+2) = (2+1+2) + (2+2+2) + (2+3+2) + (2+4+2)$

When i=2,
$\displaystyle \Sigma^4_{j=i}(2i+j+2) = (4+2+2) + (4+3+2) + (4+4+2)$

When i=3,
$\displaystyle \Sigma^4_{j=i}(2i+j+2) = (6+3+2) + (6+4+2)$

When i=4,
$\displaystyle \Sigma^4_{j=i}(2i+j+2) = (8+4+2)$

When i=5,
$\displaystyle \Sigma^4_{j=i}(2i+j+2) = 0$

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