Question

find the sum of the finite arithmetic series.
\\(\sum_{n=1}^{12} (4n - 1)\\)

Explanation:

Step1: Recall the formula for the sum of a finite arithmetic series

The sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by \( S_n=\frac{n}{2}(a_1 + a_n) \), where \( a_1 \) is the first term and \( a_n \) is the \( n \)-th term. Also, for a series defined by \( \sum_{k = 1}^{n}(ak + b) \), the first term \( a_1 \) is obtained by substituting \( k = 1 \), and the \( n \)-th term \( a_n \) is obtained by substituting \( k=n \). Here, \( n = 12 \), the general term is \( a_n=4n - 1 \).

Step2: Find the first term \( a_1 \)

Substitute \( n = 1 \) into \( 4n-1 \): \( a_1=4(1)-1=4 - 1 = 3 \).

Step3: Find the 12th term \( a_{12} \)

Substitute \( n = 12 \) into \( 4n - 1 \): \( a_{12}=4(12)-1=48 - 1 = 47 \).

Step4: Calculate the sum \( S_{12} \)

Using the formula \( S_n=\frac{n}{2}(a_1 + a_n) \), with \( n = 12 \), \( a_1 = 3 \), and \( a_{12}=47 \):
\( S_{12}=\frac{12}{2}(3 + 47) \)
First, calculate \( \frac{12}{2}=6 \), and \( 3 + 47 = 50 \). Then, \( 6\times50=300 \).

Answer:

300