Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
4. Arithmetic Series
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Exercise 47 Page 592

Practice makes perfect
a Note that the top row contains 2 cans. Moreover, each row contains 1 more can than the previous row. Therefore, the number of cans in each row form an arithmetic sequence. Let's recall the explicit rule for this type of sequence.
a_n=a+(n-1)d Here, a is the first term and d is the common difference. Let's substitute a= 2 and d= 1 into the above formula and simplify.
a_n=a+(n-1)d
a_n= 2+(n-1) 1
Simplify right-hand side
a_n=2+(n-1)
a_n=2+n-1
a_n=n+1
The explicit formula for the sequence of the number of cans is a_n=n+1.
b The related arithmetic series is the sum of the terms of the sequence from Part A. Since any triangle formed by cans has a first row, the lower limit of the summation is 1.
∑_(n= 1)^?(n+1)Let's now determine the upper limit. Note that the last row has 10 cans. To obtain n, the row that contains 10 cans, we will substitute 10 for a_n in the formula a_n=n+1.
a_n=n+1
10 = n+1
Solve for n
9 = n
n=9
The triangle has 9 rows. This is the upper limit for the summation. ∑_(n=1)^9(n+1)
c In Part A, we obtained that the number of cans in the nth row is given by the formula a_n=n+1. Therefore, to find the number of cans in the 17th row we will substitute 17 for n in this formula and simplify.
a_n=n+1
a_(17)= 17+1
a_(17)=18
There are 18 cans in the 17th row.
d Let's start by considering the formula for the sum of a finite arithmetic series.
S_n=n/2(a_1 + a_n) Here, n is the number of terms of the series and a_1 and a_n are the first and last terms, respectively. We know that the first term of our series is a_1= 2. Moreover, from Part A, we know that a_n=n+1. S_n=n/2(a_1 + a_n) ⇓ S_n=n/2( 2 + n+1) To determine whether a triangle can have 110 cans, we will substitute this number for S_n in the above formula.
S_n = n/2(2+n+1)
110 = n/2(2+n+1)
Simplify
220=n(2+n+1)
220=n(3+n)
220=3n+n^2
0=3n+n^2-220
0=n^2+3n-220
n^2+3n-220=0
Note that we have a quadratic equation with coefficients a= 1, b= 3, and c= - 220. To solve it we will use the Quadratic Formula.
n=- b± sqrt(b^2-4ac)/2a
n=- 3± sqrt(3^2-4( 1)( - 220))/2( 1)
Solve for n
n=- 3± sqrt(9-4(1)(- 220))/2(1)
n=- 3± sqrt(9-4(- 220))/2
n=- 3± sqrt(9-(- 880))/2
n=- 3± sqrt(889)/2
n≈ - 3± 29.8/2
lcn≈ - 3+ 29.82 & (I) n≈ - 3- 29.82 & (II)

(I), (II): Add and subtract terms

ln≈ 26.82 n≈ - 32.82

(I), (II): Calculate quotient

ln≈ 13.4 n≈ - 16.4
Be aware that n represents the nth row in the triangle. Therefore, a value of n that is not a natural number does not make sense. We supposed the triangle had 110 cans, and arrived to the conclusion that there are either 13.4 or - 16.4 rows. None of these numbers make sense, so the triangle cannot have 110 cans. By the same method let's see if the triangle can have 140 cans.
Cans Substitute Simplify Value of n
110 110 = n/2(2+n+1) n^2+3n-220=0 n≈ 13.4 or n≈ - 16.4
140 140 = n/2(2+n+1) n^2+3n-280=0 n≈ 15.3 or n≈ - 18.3

Since the triangle cannot have 15.3 nor - 18.3 rows, we conclude that it cannot have 140 cans.