To find the sum of the first 15 terms of the arithmetic series \(3 + 6 + 9 + \cdots\), we can use the formula for the sum of the first \(n\) terms of an arithmetic series:
[tex]\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \][/tex]
where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( n \) is the number of terms,
- \( a \) is the first term of the series,
- \( d \) is the common difference between the terms.
Given the series \( 3 + 6 + 9 + \cdots \):
- The first term \( a \) is 3.
- The common difference \( d \) is 3.
- The number of terms \( n \) is 15.
Plugging these values into the formula, we get:
[tex]\[ S_{15} = \frac{15}{2} \left(2 \times 3 + (15 - 1) \times 3\right) \][/tex]
Let's break it down step by step:
1. Calculate \( 2a \):
[tex]\[ 2 \times 3 = 6 \][/tex]
2. Calculate \( (n - 1) \):
[tex]\[ 15 - 1 = 14 \][/tex]
3. Calculate \( (n - 1) \times d \):
[tex]\[ 14 \times 3 = 42 \][/tex]
4. Add \( 2a \) and \( (n - 1) \times d \):
[tex]\[ 6 + 42 = 48 \][/tex]
5. Multiply by \( \frac{n}{2} \):
[tex]\[ \frac{15}{2} \times 48 = \frac{15 \times 48}{2} = \frac{720}{2} = 360 \][/tex]
Therefore, the sum of the first 15 terms of the series is:
[tex]\[ S_{15} = 360 \][/tex]
