S_n=(n/2)*(a_ 1+a_n), where s _ n represents the sum of the first n terms, n represents the number of terms, a_ 1 represents the first term, and a_n represents the last term.
1. Definition of arithmetic series
Arithmetic progression refers to a series in which the difference between each term and its previous term is equal. The letter A is commonly used for the first item, D for the tolerance, N for the number of items, and an for the nth item.
2. Definition of the sum of the first n items
The sum of the first n terms represents the sum of the first n terms in the sequence, which is represented by Sn.
3. Derive the summation formula
Let's deduce arithmetic progression's first n terms and formulas. First of all, we use mathematical expressions to represent arithmetic progression:
A 1, A 1+D, A 1+2D, ..., A 1+(n- 1) D, where A 1 represents the first term and d represents the tolerance. Then, the first n terms and Sn are the sum of each term in the sequence, that is, Sn = a1+(a1+d)+(a1+2d)+...+[a1+(n-/kloc-0).
Now we arrange arithmetic progression in reverse order, and get: an, an-d, an-2d, ..., an-(n- 1) d, where an stands for the last term. Since the sum of the first n terms in arithmetic progression is equal to the sum of the first n terms in reverse order, we can add these two terms, namely: Sn+Sn = (a1+an)+(a1+d+an-d)+(a1+2d+an-2d). ...
In the above formula, the left is the sum of two SNs, and the right is the result of adding the corresponding items. We can observe that the sum of the numbers in each pair of brackets is equal to an+a 1. The equation on the right has n parentheses, which can be simplified as 2Sn=n(an+a 1). Finally, divide both sides of the above formula by 2 at the same time to get the first n terms of arithmetic progression and the formula: Sn=(n/2)(an+a 1).
4. Application example
Give an example to illustrate how to apply this formula. Suppose there is a arithmetic progression: 3,6,9, 12, 15, and find the sum of the first four terms. The first term a 13, the tolerance d=6-3=3, the number of terms n=4, and the last term an=3+(4- 1)*3= 12.
According to the formula: sn = (n/2) (an+a1) = (4/2) (12+3) = 2 *15 = 30, so the sum of the first four terms of this arithmetic progression is 30.
Summary: By deducing the summation formula, we can efficiently calculate the sum of the first n terms of arithmetic progression. This formula has many applications in mathematics and practice, such as arithmetic progression summation in finance, engineering, statistics and other fields.