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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.
As we can see, each sequence has a common difference, implying that the sequences are arithmetic.
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
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.
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)
S_n = n/2(a_1+a_n)
n= 5
a_1= 5, a_5= 21
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The second series is ∑ _(n=1)^5(2n-1), which tells us that b_1 = 1 and b_5=9.
n= 5
b_1= 1, b_5= 9
Add terms
Multiply
Calculate quotient