Addition Tricks (3)

The series solution given is probably the fastest way to calculate the sum. However, since this is a Level 1 Algebra question, I would like to share a method that does not require knowledge on series.

Now, we are looking for the sum below:

$1+2+3+4+5+\dots+996+997+998+999+1000$

Notice that:

$1+1000 = 1001\\2+999 = 1001\\3+998 = 1001\\4+997 = 1001\\5+996=1001\\\dots\\\dots\\\dots\\500+501=1001$

The first and last number adds up to $1001$ , the second and second last number also adds up to $1001$ , and the third and third last number also adds up to $1001$ . This pattern continues until we reach the end, where $500+501 = 1001$ . From this list, we can know that the sum actually consists of $1001$ added $500$ times. Therefore,

$1+2+3+4+5+\dots+996+997+998+999+1000\\=500 \times 1001 = \boxed{500500}$

It forms an Arithmetic Progression with a common difference of $1$ .

$S = \frac{n}{2}(a_1 + a_n) = \frac{1000}{2}(1 + 1000) = 500500$

$\Large \displaystyle \sum_{n=1}^{x} = \dfrac{n (n+1)}{2}$
$\therefore \displaystyle \sum_{n=1}^{1000} = \dfrac{1000 × 1001}{2}$
= $500 × 1001 = \boxed{500500}$