To determine the [tex]\(n\)[/tex]-th term of the sequence [tex]\((a_n)\)[/tex] from the given expression for the partial sum [tex]\(\sum_{k=1}^n (6k + 2)\)[/tex], we need to understand how the partial sum relates to the individual terms of the sequence.
The partial sum of the sequence [tex]\(a_n\)[/tex] is given as:
[tex]\[
S_n = \sum_{k=1}^n (6k + 2)
\][/tex]
If we denote the [tex]\(n\)[/tex]-th term of the sequence by [tex]\(a_n\)[/tex], then the [tex]\(n\)[/tex]-th partial sum [tex]\(S_n\)[/tex] is the sum of the first [tex]\(n\)[/tex] terms of the sequence:
[tex]\[
S_n = a_1 + a_2 + a_3 + \ldots + a_n
\][/tex]
To find the [tex]\(n\)[/tex]-th term [tex]\(a_n\)[/tex], we need to express [tex]\(a_n\)[/tex] in terms of [tex]\(S_n\)[/tex] and the previous partial sum [tex]\(S_{n-1}\)[/tex]:
[tex]\[
a_n = S_n - S_{n-1}
\][/tex]
Given [tex]\(S_n = \sum_{k=1}^n (6k + 2)\)[/tex], we can examine this more closely.
First, calculate [tex]\(S_n\)[/tex] explicitly:
[tex]\[
S_n = \sum_{k=1}^n (6k + 2) = 6 \sum_{k=1}^n k + \sum_{k=1}^n 2
\][/tex]
We know the sum of the first [tex]\(n\)[/tex] natural numbers is given by:
[tex]\[
\sum_{k=1}^n k = \frac{n(n+1)}{2}
\][/tex]
The sum of [tex]\(n\)[/tex] repetitions of the constant 2 is:
[tex]\[
\sum_{k=1}^n 2 = 2n
\][/tex]
Thus:
[tex]\[
S_n = 6 \left(\frac{n(n+1)}{2}\right) + 2n = 3n(n+1) + 2n = 3n^2 + 3n + 2n = 3n^2 + 5n
\][/tex]
So:
[tex]\[
S_n = 3n^2 + 5n
\][/tex]
Next, calculate [tex]\(S_{n-1}\)[/tex]:
[tex]\[
S_{n-1} = 3(n-1)^2 + 5(n-1)
\][/tex]
[tex]\[
S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2
\][/tex]
Then find [tex]\(a_n\)[/tex] by subtracting [tex]\(S_{n-1}\)[/tex] from [tex]\(S_n\)[/tex]:
[tex]\[
a_n = S_n - S_{n-1} = (3n^2 + 5n) - (3n^2 - n - 2)
\][/tex]
[tex]\[
a_n = 3n^2 + 5n - 3n^2 + n + 2
\][/tex]
[tex]\[
a_n = 6n + 2
\][/tex]
Therefore, the [tex]\(n\)[/tex]-th term of the sequence is:
\[
\boxed{6n + 2}
\
