Write the arithmetic series using summation notation. ABC 41. 43. 45. 12+9+6+3+0+(-3) 48 +60 + 72 +84 +96 -1 + (-13) + (-25) + (-37) + (-49) + ... Choices for 41-45: Α) Σ36 +12k E=1 E) Σ36+12k k=1 ∞ ΑΕ) Σ12–3 k=1 5 CD) Σ11–12k F-1 ABD) Σ48 – 12k +1 Β) Σ-1-12k 6-1 ΑΒ) Σ1.7+1. k=1 ζ BC) Σ48 – 12k k=1 5 CE) Σ-10 + 3k *1 ΑΒΕ) Σ ~10 + 3k k=1 6 k=1 42. ο Σ12–3k 44. k=1 AC) Σ11–12k -10 + (-7) + (-4) + (-1) + 2 + ... 2.8 +3.9+ 5.0+ 6.1 +7.2 X BD) Σ-13 + 3k k−1 x DE) Σ2.8 + 1.1k k=1 ACD) Σ−1−12k k=1 D) Σ2.8 +1.1k k=1 5 AD) Σ−13 + 3k k=1 5 ΒΕ) Σ1.7+1.1k k=1 6 ABC) Σ15–3k k=1 ACE) Σ15–3k k=1

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter9: Sequences, Probability And Counting Theory
Section: Chapter Questions
Problem 23RE: Use the formula for the sum of the first ii terms of an arithmetic series to find the sum of the...
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Question
42,43,44,45
Write the arithmetic series using summation notation.
ABC
12+9+6+3+0+(-3)
41.
43.
45.
48 +60 + 72 +84 +96
-1 + (-13) + (-25) + (-37) + (-49) + ...
Choices for 41-45:
Α) Σ36 +12k
k=1
E) Σ36+12k
k=1
ΑΕ) Σ12–3
k−1
5
CD) Σ11–12k
k 1
ABD) Σ48 – 12k
+1
Β) Σ-1-12k
k−1
ΑΒ) Σ1.7+1.
k=1
ζ
BC) Σ48 – 12k
k=1
5
CE) Σ-10 + 3k
* 1
ΑΒΕ) Σ –10 + 3k
k=1
6
42.
k=1
44.
c) Σ12–3k
-10 + (-7) + (-4) + (-1) + 2 + ...
2.8 +3.9+ 5.0+ 6.1 +7.2
AC) Σ11 – 12k
k=1
BD) Σ-13 + 3k
k=1
DE) Σ2.8 + 1.1k
k=1
ACD) Σ−1−12k
k=1
D) Σ2.8 +1.1k
k=1
5
AD) Σ−13 + 3k
k=1
5
ΒΕ) Σ1.7+1.1k
k=1
6
ABC) Σ15–3k
k-1
ACE) Σ15–3k
k=1
Transcribed Image Text:Write the arithmetic series using summation notation. ABC 12+9+6+3+0+(-3) 41. 43. 45. 48 +60 + 72 +84 +96 -1 + (-13) + (-25) + (-37) + (-49) + ... Choices for 41-45: Α) Σ36 +12k k=1 E) Σ36+12k k=1 ΑΕ) Σ12–3 k−1 5 CD) Σ11–12k k 1 ABD) Σ48 – 12k +1 Β) Σ-1-12k k−1 ΑΒ) Σ1.7+1. k=1 ζ BC) Σ48 – 12k k=1 5 CE) Σ-10 + 3k * 1 ΑΒΕ) Σ –10 + 3k k=1 6 42. k=1 44. c) Σ12–3k -10 + (-7) + (-4) + (-1) + 2 + ... 2.8 +3.9+ 5.0+ 6.1 +7.2 AC) Σ11 – 12k k=1 BD) Σ-13 + 3k k=1 DE) Σ2.8 + 1.1k k=1 ACD) Σ−1−12k k=1 D) Σ2.8 +1.1k k=1 5 AD) Σ−13 + 3k k=1 5 ΒΕ) Σ1.7+1.1k k=1 6 ABC) Σ15–3k k-1 ACE) Σ15–3k k=1
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