Sure, let's derive the formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series step-by-step.
### Step-by-Step Derivation:
#### Step 1: Define the Arithmetic Series
An arithmetic series is formed by adding the terms of an arithmetic sequence. An arithmetic sequence is defined by its first term [tex]\( a \)[/tex] and the common difference [tex]\( d \)[/tex]. The [tex]\( n \)[/tex]-th term of this sequence, [tex]\( a_n \)[/tex], can be expressed as:
[tex]\[ a_n = a + (n-1)d \][/tex]
#### Step 2: Write the Formula for the Sum of the First [tex]\( n \)[/tex] Terms
The sum of the first [tex]\( n \)[/tex] terms of the arithmetic series is denoted by [tex]\( S_n \)[/tex]. It can be written as:
[tex]\[ S_n = a + (a+d) + (a+2d) + \ldots + [a + (n-1)d] \][/tex]
#### Step 3: Write the Sum in Reverse Order
Now, write the sum [tex]\( S_n \)[/tex] in reverse order:
[tex]\[ S_n = [a + (n-1)d] + [a + (n-2)d] + \ldots + (a+d) + a \][/tex]
#### Step 4: Add the Two Equations
Add the original sum and the reversed sum term by term:
[tex]\[ S_n + S_n = (a + [a + (n-1)d]) + ([a+d] + [a + (n-2)d]) + \ldots + ([a + (n-1)d] + a) \][/tex]
Simplify the terms inside the parentheses:
[tex]\[ S_n + S_n = 2S_n = [2a + (n-1)d] + [2a + (n-1)d] + \ldots + [2a + (n-1)d] \][/tex]
Notice that there are [tex]\( n \)[/tex] terms of [tex]\([2a + (n-1)d]\)[/tex] in the equation above.
#### Step 5: Simplify the Expression
Factor [tex]\( n \)[/tex] out of the right-hand side:
[tex]\[ 2S_n = n[2a + (n-1)d] \][/tex]
#### Step 6: Isolate [tex]\( S_n \)[/tex]
To find [tex]\( S_n \)[/tex], divide both sides by 2:
[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]
### Conclusion
We have derived the formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic series:
[tex]\[ S_n = \frac{1}{2} n [2a + (n-1)d] \][/tex]
This completes the proof.
