Question

question 7 of 10
if the nth partial sum of a sequence ( a_n ) is given by
( sum_{k=1}^{n} 2k + 4 )
then what is the ( n^{\text{th}} ) term of the sequence?

a. 6

b. ( 2n + 4 )

c. ( 6 + 8 + 10 + 12 + dots + (2n + 4) )

d. ( 2(1) + 4 + 2(n) + 4 )

Explanation:

Step1: Recall the formula for the nth term of a sequence in terms of partial sums

The nth term \( a_n \) of a sequence can be found using the relationship between the nth partial sum \( S_n \) and the (n - 1)th partial sum \( S_{n-1} \), which is \( a_n=S_n - S_{n - 1} \).

First, let's write the formula for \( S_n \) and \( S_{n-1} \).

Given \( S_n=\sum_{k = 1}^{n}(2k + 4) \), then \( S_{n-1}=\sum_{k=1}^{n - 1}(2k+4) \)

Step2: Calculate \( a_n=S_n - S_{n-1} \)

\( S_n=\sum_{k = 1}^{n}(2k + 4)=2\sum_{k = 1}^{n}k+\sum_{k = 1}^{n}4 \)

We know that \( \sum_{k = 1}^{n}k=\frac{n(n + 1)}{2} \) and \( \sum_{k = 1}^{n}4=4n \)

So \( S_n=2\times\frac{n(n + 1)}{2}+4n=n(n + 1)+4n=n^2+n + 4n=n^2+5n \)

Similarly, \( S_{n-1}=\sum_{k = 1}^{n - 1}(2k + 4)=2\sum_{k = 1}^{n - 1}k+\sum_{k = 1}^{n - 1}4 \)

\( \sum_{k = 1}^{n - 1}k=\frac{(n - 1)n}{2} \) and \( \sum_{k = 1}^{n - 1}4=4(n - 1) \)

So \( S_{n-1}=2\times\frac{(n - 1)n}{2}+4(n - 1)=(n - 1)n+4n-4=n^2 - n+4n - 4=n^2+3n - 4 \)

Now, \( a_n=S_n - S_{n-1}=(n^2+5n)-(n^2+3n - 4)=n^2+5n - n^2-3n + 4=2n + 4 \)

Alternatively, we can think about the nth term of the sum \( \sum_{k = 1}^{n}(2k + 4) \). The nth term of the sequence whose partial sum is this sum is just the term when \( k = n \) in the summand (because the partial sum \( S_n \) is the sum of the first n terms, so the nth term \( a_n \) is the term added at the nth step, which is \( 2n+4 \)).

Answer:

B. \( 2n + 4 \)