Answer :
To determine the [tex]\( n \)[/tex]th term of a sequence given the [tex]\( n \)[/tex]th partial sum, we need to find the difference between the [tex]\( n \)[/tex]th partial sum and the [tex]\( (n-1) \)[/tex]th partial sum.
Given the [tex]\( n \)[/tex]th partial sum [tex]\( S_n \)[/tex]:
[tex]\[ S_n = \sum_{k=1}^n (2k + 4) \][/tex]
We know that the sequence term [tex]\( a_n \)[/tex] is the difference between the [tex]\( n \)[/tex]th partial sum and the [tex]\( (n-1) \)[/tex]th partial sum. In formula terms:
[tex]\[ a_n = S_n - S_{n-1} \][/tex]
First, we need to express [tex]\( S_n \)[/tex] and [tex]\( S_{n-1} \)[/tex] explicitly:
The partial sum [tex]\( S_n \)[/tex] can be calculated as follows:
[tex]\[ S_n = \sum_{k=1}^n (2k + 4) \][/tex]
This can be simplified by separating the sum into two parts:
[tex]\[ S_n = \sum_{k=1}^n 2k + \sum_{k=1}^n 4 \][/tex]
The first part is the sum of the first [tex]\( n \)[/tex] terms of 2 times [tex]\( k \)[/tex]:
[tex]\[ \sum_{k=1}^n 2k = 2 \sum_{k=1}^n k = 2 \left(\frac{n(n+1)}{2}\right) = n(n+1) \][/tex]
The second part is the sum of 4 added [tex]\( n \)[/tex] times:
[tex]\[ \sum_{k=1}^n 4 = 4n \][/tex]
Therefore,
[tex]\[ S_n = n(n + 1) + 4n = n^2 + n + 4n = n^2 + 5n \][/tex]
Next, we calculate [tex]\( S_{n-1} \)[/tex]:
[tex]\[ S_{n-1} = (n-1)^2 + 5(n-1) \][/tex]
Expanding and simplifying:
[tex]\[ S_{n-1} = n^2 - 2n + 1 + 5n - 5 = n^2 + 3n - 4 \][/tex]
Now, we find [tex]\( a_n \)[/tex]:
[tex]\[ a_n = S_n - S_{n-1} \][/tex]
Substitute [tex]\( S_n \)[/tex] and [tex]\( S_{n-1} \)[/tex]:
[tex]\[ a_n = (n^2 + 5n) - (n^2 + 3n - 4) \][/tex]
Simplify the expression:
[tex]\[ a_n = n^2 + 5n - n^2 - 3n + 4 = 2n + 4 \][/tex]
Thus, the [tex]\( n \)[/tex]th term of the sequence is:
[tex]\[ 2n + 4 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{C. \, 2n + 4} \][/tex]
Given the [tex]\( n \)[/tex]th partial sum [tex]\( S_n \)[/tex]:
[tex]\[ S_n = \sum_{k=1}^n (2k + 4) \][/tex]
We know that the sequence term [tex]\( a_n \)[/tex] is the difference between the [tex]\( n \)[/tex]th partial sum and the [tex]\( (n-1) \)[/tex]th partial sum. In formula terms:
[tex]\[ a_n = S_n - S_{n-1} \][/tex]
First, we need to express [tex]\( S_n \)[/tex] and [tex]\( S_{n-1} \)[/tex] explicitly:
The partial sum [tex]\( S_n \)[/tex] can be calculated as follows:
[tex]\[ S_n = \sum_{k=1}^n (2k + 4) \][/tex]
This can be simplified by separating the sum into two parts:
[tex]\[ S_n = \sum_{k=1}^n 2k + \sum_{k=1}^n 4 \][/tex]
The first part is the sum of the first [tex]\( n \)[/tex] terms of 2 times [tex]\( k \)[/tex]:
[tex]\[ \sum_{k=1}^n 2k = 2 \sum_{k=1}^n k = 2 \left(\frac{n(n+1)}{2}\right) = n(n+1) \][/tex]
The second part is the sum of 4 added [tex]\( n \)[/tex] times:
[tex]\[ \sum_{k=1}^n 4 = 4n \][/tex]
Therefore,
[tex]\[ S_n = n(n + 1) + 4n = n^2 + n + 4n = n^2 + 5n \][/tex]
Next, we calculate [tex]\( S_{n-1} \)[/tex]:
[tex]\[ S_{n-1} = (n-1)^2 + 5(n-1) \][/tex]
Expanding and simplifying:
[tex]\[ S_{n-1} = n^2 - 2n + 1 + 5n - 5 = n^2 + 3n - 4 \][/tex]
Now, we find [tex]\( a_n \)[/tex]:
[tex]\[ a_n = S_n - S_{n-1} \][/tex]
Substitute [tex]\( S_n \)[/tex] and [tex]\( S_{n-1} \)[/tex]:
[tex]\[ a_n = (n^2 + 5n) - (n^2 + 3n - 4) \][/tex]
Simplify the expression:
[tex]\[ a_n = n^2 + 5n - n^2 - 3n + 4 = 2n + 4 \][/tex]
Thus, the [tex]\( n \)[/tex]th term of the sequence is:
[tex]\[ 2n + 4 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{C. \, 2n + 4} \][/tex]
