# How do you find the sum of the first 1000 positive even integers?

Oct 16, 2016

An even integer number $n$ can be written as $n = 2 \cdot k$, with $k$ an integer

#### Explanation:

The sum you are looking for is:

$2 + 4 + 6 + 8 + \ldots + 2000$, but taking 2 as a common factor this is the same as:

$2 \left(1 + 2 + 3 + 4 + \ldots + 1000\right)$

Now we can use the formula for the sum:

${\sum}_{n = 1}^{1000} n = \frac{1000 \cdot 1001}{2} = 500 \cdot 1001 = 500500$

Using this now we have:

$2 \left(1 + 2 + 3 + 4 + \ldots + 1000\right) = 2 \cdot {\sum}_{n = 1}^{1000} n = \frac{1000 \cdot 1001}{2} =$

$= 2 \cdot 500500 = 1001000$