# Double Summation with a constant as the Summand

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:

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

$$n^2 + n$$
• The answer given to you fails for $n=2$. – Git Gud Aug 31 '13 at 18:14
• Are you sure you've read the limits correctly in the original question - check that the upper limit of the inner sum is not $i$. – Mark Bennet Aug 31 '13 at 19:16