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$\begingroup$

$\sum_{i=0}^{n} \sum_{j=0}^{i-1} (i+1)$

Not totally sure how to get passed this one. I brought it down from a triple to a double, but I'm having trouble with the fact that $i$ is the what is being added.

My thought is that it will be $\sum_{i=0}^{n} (i+1)^2$ but I cant say that with total certainty.

Thanks for the help.

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    @J.M. I'm aware that is the convention. "Meaningless" was a poor choice of wording on my part. My intention was just to ensure that there wasn't a typo or anything mixed in there.2012-02-11

1 Answers 1

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$\sum_{i=0}^{n} \sum_{j=0}^{i-1} (i+1)$

is composed of tow sums. Let's figure the inner sum first:

$\sum_{j=0}^{i-1} (i+1) = (i+1)(i-1-0+1) = (i+1)(i)$

Now you can evaluate the outer sum:

$\sum_{i=0}^{n} (i+1)(i)$ which is: $\sum_{i=0}^{n} i^2+i$

You can split the summs as:

$\sum_{i=0}^{n} i^2$ + $\sum_{i=0}^{n} i$

The following link may help you further: Some summations of polynomial expressions

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    Yes, I edited to post to show that.2012-02-11