The objective is to find a formula to calculate the sum of the first terms of an arithmetical progression without any need to calculate them.

Let's consider the arithmetical progression: $$a_n=3n-1$$. Let's consider only the first six terms: $$$a_n=(2,5,8,11,14,17,\ldots)$$$ We represent them in a grid, with the same numbers placed in an inverted way:

The sum of the first six terms is the area bounded by the red polygon, which by construction is the same area as the white polygon, and both are half the area of the entire rectangle.

The fact of obtaining a rectangle shows us an important property of the arithmetical progressions. We can see that the base of the rectangle has a length equal to the sum of the first and the sixth terms, and this value is the same as the sum of the fifth and the second term, and the sum of the third and the forth; and these three pairs of terms are equidistant from the extremes, i.e., the first and sixth.

In general, we can say that if we consider $$n$$ terms of an arithmetical progression, the sum of two equidistant terms from the extremes is the same as the sum of the extremes (the sum of the first and the last term is the same as the sum of the second and the penultimate term, and the sum of the third and the antepenultimate term and so on, no matter how many terms we are considering in an arithmetical progression).

Going back to the previous progression, if we look at the rectangle and calculate its area, its base, as we have already seen, is the sum of the first and the last terms: $$a_1+a_6$$, and the height is equal to the quantity of numbers that we want to add: $$6$$. This way the area of the rectangle is:

$$$A_{rectangle}=(a_1+a_6)\cdot 6$$$

Therefore, the sum of the first six terms is: $$$S_6=\dfrac{A_{rect.}}{2}=\dfrac{6\cdot(a_1+a_6)}{2}=\dfrac{6\cdot(2+17)}{2}=57$$$

In general, the sum of $$n$$ terms of an arithmetical progression is the semiproduct of the number of terms by the sum of the extremes:

$$$S_n=\dfrac{n\cdot(a_1+a_n)}{2}$$$

For example, we want to calculate the sum of the first thousand natural numbers which are multiples of five.

The first natural multiples of five are: $$0,5,10,15,20,\ldots$$

Let's observe that they form an arithmetical progression of difference $$d=5$$ and the first term is $$a_1=0$$. Then, the term that occupies the position a thousand is $$$a_{1000}=999\cdot5=4995$$$ Applying the formula we obtain $$$S_{1000}=\dfrac{1000\cdot(0+4995)}{2}=2.477.500$$$

The sum of the $$1000$$ first multiples of $$5$$ is $$2.477.500$$

To make the writing easier and to simplify the notation if we are working with a large amount of numbers that we cannot write explicitly, to denote the sum we will use the Greek capital letter Sigma: $$\sum$$.

We will write the variable we are adding and the initial term on the bottom and the last term to be added on the top. Next to the letter sigma, we will write the general term of the progression we want to add up.

In the previous example, we will add the first thousand multiples of five as $$$S_{1000}=\sum_{n=1}^{1000} 5(n-1)$$$

And adding the first six terms of the succession $$a_n=3n-1$$ can be written as: $$$S_{6}=\sum_{n=1}^{6} 3n-1 $$$