Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
4. Arithmetic Series
Continue to next subchapter

Exercise 50 Page 592

Practice makes perfect
a Let's consider the following two explicit formulas.


a_n = 4n+1 and b_n = 2n-1

Since each function is a linear function of n, we have that each formula represents an arithmetic sequence. Keep in mind that this is just an example and your answer may vary.

Extra

Verifying the Sequences are Arithmetic

Let's write the first three terms of each sequence and find the difference between consecutive terms.

As we can see, each sequence has a common difference, implying that the sequences are arithmetic.

b In the previous part we wrote the following two arithmetic sequences.
a_n = 4n+1 and b_n = 2n-1 Before finding the first five terms of each related series, we will find the first five terms of each sequence. To do this we substitute n=1, 2, 3, 4, and 5.
n a_n = 4n+1 b_n = 2n-1
1 a_1 = 4( 1)+1=5 b_1 = 2( 1)-1 = 1
2 a_2 = 4( 2)+1 = 9 b_2 = 2( 2)-1 = 3
3 a_3 = 4( 3)+1 = 13 b_3 = 2( 3)-1 = 5
4 a_4 = 4( 4)+1 = 17 b_4 = 2( 4)-1 = 7
5 a_5 = 4( 5)+1 = 21 b_5 = 2( 5)-1 = 9

The first five terms of the related series for a_n are shown below.


5+9+13+17+21

Similarly, the first five terms of the related series for b_n are the following.


1+3+5+7+9

c In the previous part we wrote the following two series.
Series 1:& 5+9+13+17+21 Series 2:& 1+3+5+7+9 A series written in summation notation has the form ∑ _(n=a)^ba_n, where a_n is the explicit formula, a is the lower limit, and b is the upper limit. In the two series written above there are five terms.

a=1 and b=5 For the first series, the explicit formula is a_n =4n+1. With this information, let's write it in summation notation.


5+9+13+17+21 = ∑ _(n= 1)^5 (4n+1)

Similarly, the explicit formula for the second series is b_n = 2n-1, which leads us to the following formula.


1+3+5+7+9 = ∑ _(n= 1)^5 (2n-1)

d To evaluate a finite series means to compute S_n. That is the sum of the first n terms. To do this, we apply the formula to find the sum of a finite arithmetic series.
S_n = n/2(a_1+a_n)In our case both series have 5 terms, which means that n=5. The first series is ∑ _(n=1)^5(4n+1), which means that a_1=5 and a_5 = 21.
S_n = n/2(a_1+a_n)
S_5 = 5/2(a_1+a_5)
Simplify right-hand side
S_5 = 5/2( 5+ 21)
S_5 = 5/2(26)
S_5 = 130/2
S_5 = 65
The second series is ∑ _(n=1)^5(2n-1), which tells us that b_1 = 1 and b_5=9.
S_n = n/2(b_1+b_n)
S_5 = 5/2(b_1+b_5)
Simplify right-hand side
S_5 = 5/2( 1+ 9)
S_5 = 5/2(10)
S_5 = 50/2
S_5 = 25