Answer :
To find the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex] of the sequence where the [tex]\( n \)[/tex]-th partial sum is given by
[tex]\[ S_n = \sum_{k=1}^n \left( 6k + 2 \right), \][/tex]
we need to analyze the given sum.
1. Understand the given partial sum expression:
[tex]\[ S_n = \sum_{k=1}^n \left( 6k + 2 \right). \][/tex]
2. Simplify the sum:
Let's break down the sum by splitting it into two parts:
[tex]\[ S_n = \sum_{k=1}^n 6k + \sum_{k=1}^n 2. \][/tex]
3. Evaluate each part of the sum separately:
- The first part [tex]\( \sum_{k=1}^n 6k \)[/tex] is a sum of an arithmetic series:
[tex]\[ \sum_{k=1}^n 6k = 6 \sum_{k=1}^n k = 6 \cdot \frac{n(n+1)}{2} = 3n(n+1). \][/tex]
- The second part [tex]\( \sum_{k=1}^n 2 \)[/tex] is simply [tex]\( 2 \)[/tex] added [tex]\( n \)[/tex] times:
[tex]\[ \sum_{k=1}^n 2 = 2n. \][/tex]
4. Combine the results:
[tex]\[ S_n = 3n(n+1) + 2n. \][/tex]
5. Simplify the combined expression:
[tex]\[ S_n = 3n^2 + 3n + 2n = 3n^2 + 5n. \][/tex]
6. Find the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex]:
The [tex]\( n \)[/tex]-th term of the sequence [tex]\( a_n \)[/tex] is given by [tex]\( S_n - S_{n-1} \)[/tex]:
[tex]\[ a_n = S_n - S_{n-1}. \][/tex]
7. Express [tex]\( S_{n-1} \)[/tex]:
Substitute [tex]\( n-1 \)[/tex] into the partial sum formula:
[tex]\[ S_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2. \][/tex]
8. Calculate the difference [tex]\( S_n - S_{n-1} \)[/tex]:
[tex]\[ a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 3n^2 + 5n - 3n^2 + n + 2 = 6n + 2. \][/tex]
Therefore, the [tex]\( n \)[/tex]-th term of the sequence [tex]\( a_n \)[/tex] is
[tex]\[ \boxed{6n + 2}. \][/tex]
[tex]\[ S_n = \sum_{k=1}^n \left( 6k + 2 \right), \][/tex]
we need to analyze the given sum.
1. Understand the given partial sum expression:
[tex]\[ S_n = \sum_{k=1}^n \left( 6k + 2 \right). \][/tex]
2. Simplify the sum:
Let's break down the sum by splitting it into two parts:
[tex]\[ S_n = \sum_{k=1}^n 6k + \sum_{k=1}^n 2. \][/tex]
3. Evaluate each part of the sum separately:
- The first part [tex]\( \sum_{k=1}^n 6k \)[/tex] is a sum of an arithmetic series:
[tex]\[ \sum_{k=1}^n 6k = 6 \sum_{k=1}^n k = 6 \cdot \frac{n(n+1)}{2} = 3n(n+1). \][/tex]
- The second part [tex]\( \sum_{k=1}^n 2 \)[/tex] is simply [tex]\( 2 \)[/tex] added [tex]\( n \)[/tex] times:
[tex]\[ \sum_{k=1}^n 2 = 2n. \][/tex]
4. Combine the results:
[tex]\[ S_n = 3n(n+1) + 2n. \][/tex]
5. Simplify the combined expression:
[tex]\[ S_n = 3n^2 + 3n + 2n = 3n^2 + 5n. \][/tex]
6. Find the [tex]\( n \)[/tex]-th term [tex]\( a_n \)[/tex]:
The [tex]\( n \)[/tex]-th term of the sequence [tex]\( a_n \)[/tex] is given by [tex]\( S_n - S_{n-1} \)[/tex]:
[tex]\[ a_n = S_n - S_{n-1}. \][/tex]
7. Express [tex]\( S_{n-1} \)[/tex]:
Substitute [tex]\( n-1 \)[/tex] into the partial sum formula:
[tex]\[ S_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2. \][/tex]
8. Calculate the difference [tex]\( S_n - S_{n-1} \)[/tex]:
[tex]\[ a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 3n^2 + 5n - 3n^2 + n + 2 = 6n + 2. \][/tex]
Therefore, the [tex]\( n \)[/tex]-th term of the sequence [tex]\( a_n \)[/tex] is
[tex]\[ \boxed{6n + 2}. \][/tex]
