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As part of the preparation for an Exam, I'm working through a simple double sum exercise that has a constant as the Summand. The exercise has the answer (result) available but not the resolution. I attempted to solve it, but apparently I've reached a wrong answer.

The question:

Calculate the following sum: $$\sum_{i = 1}^n \sum_{j = 1}^n 2$$

My attempted resolution:

$$\sum_{i = 1}^n \sum_{j = 1}^n 2 = \sum_{i = 1}^n \underbrace{(2 + 2 + \cdots + 2)}_{\text{n times}} = \sum_{i = 1}^n 2n = 2n \sum_{i = 1}^n 1 = 2n \times \underbrace{(1 + 1 + \cdots + 1)}_{\text{n times}} = 2n \times n = 2n^2$$

The right answer:

$$n^2 + n$$

What did I do wrong? How should I have solved this?

EDIT: Mistery solved! I had made a small but important mistake when I copied the question from the paper it was written. The lower limit of the inner sum is "j=i" ("j equals lowercase i") instead of "j=1" ("j equals one"). Thanks to everyone who commented (T.Bongers, Git Gud and, particularly, Mark Bennet who correctly guessed that I might have made a mistake in one of the limits).

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    $\begingroup$ No, you've done it correctly. There appears to be a mistake in the answer key. $\endgroup$ – user61527 Aug 31 '13 at 18:12
  • $\begingroup$ The answer given to you fails for $n=2$. $\endgroup$ – Git Gud Aug 31 '13 at 18:14
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    $\begingroup$ Are you sure you've read the limits correctly in the original question - check that the upper limit of the inner sum is not $i$. $\endgroup$ – Mark Bennet Aug 31 '13 at 19:16
  • $\begingroup$ @MarkBennet Thanks for the tip! Actually, the upper limit of both sums is "n" (as I wrote), but the LOWER limit of the inner sum is "i" instead of 1 (so it's "j=i" instead of "j=1"). Good catch! $\endgroup$ – ricmarques Aug 31 '13 at 20:16
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    $\begingroup$ @ricmarques That's cunning of the examiner - but you'll be looking out for it come the real thing - which is the point of practice. One more trick you'll see through! $\endgroup$ – Mark Bennet Aug 31 '13 at 20:24

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