Write the arithmetic series using summation notation. 41. 12 +9+6+3+0+(-3) 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 k-1 Β) Σ−1−12k k=1 ΑΒ) Σ1.7+1.1k k=1 BC) Σ48 – 12k k=1 5 CE) Σ-10 + 3k k=1 ΑΒΕ) Σ-10 + 3k 53 42. k=1 44. 6 ο Σ12–3k Στην k=1 -10 + (-7)+(-4) + (-1) + 2 + ... 2.8 +3.9+ 5.0+ 6.1 +7.2 AC) Σ11-12k S 00 BD) Σ−13 + 3k k=1 ACD) Σ-1 DE) Σ2.8 +1.1k k=1 –1-12k D) Σ2.8 +1.1k k=1 5 AD) Σ−13 + 3k k=1 5 ΒΕ) Σ1.7+1.1k ABC) Σ15–3k k=1 k=1 ACE) Σ15–3k k=1
Write the arithmetic series using summation notation. 41. 12 +9+6+3+0+(-3) 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 k-1 Β) Σ−1−12k k=1 ΑΒ) Σ1.7+1.1k k=1 BC) Σ48 – 12k k=1 5 CE) Σ-10 + 3k k=1 ΑΒΕ) Σ-10 + 3k 53 42. k=1 44. 6 ο Σ12–3k Στην k=1 -10 + (-7)+(-4) + (-1) + 2 + ... 2.8 +3.9+ 5.0+ 6.1 +7.2 AC) Σ11-12k S 00 BD) Σ−13 + 3k k=1 ACD) Σ-1 DE) Σ2.8 +1.1k k=1 –1-12k D) Σ2.8 +1.1k k=1 5 AD) Σ−13 + 3k k=1 5 ΒΕ) Σ1.7+1.1k ABC) Σ15–3k k=1 k=1 ACE) Σ15–3k k=1
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
Transcribed Image Text:Write the arithmetic series using summation notation.
41. 12 +9+6+3+0+(-3)
43.
45.
48 +60 + 72 +84 +96
-1 + (-13) + (-25) + (-37) + (-49) + ...
Choices for 41-45:
Α) Σ36+12k
k=1
5
E) Σ36+12k
k=1
ΑΕ) Σ12–3k
k=1
5
cD) Σ11-12k
re
k=1
ABD) Σ48–12k
k=1
Β) Σ−1−12k
k=1
ΑΒ) Σ1.7+1.1
k=1
5
BC) Σ48 – 12k
k=1
CE) Σ-
k=1
–10 + 3k
ΑΒΕ) Σ−10 + 3k
k=1
k=1
42.
44.
Ε
c) Σ12–3k
k=1
n
AC) Σ11–12k
-10 + (-7)+(-4) + (-1) + 2 + ...
2.8 +3.9+ 5.0+ 6.1 +7.2
4701
BD) Σ-13 + 3k
k=1
DE) 2.8.
k=1
2.8+1.1k
k=1
AcD) Σ−1−12k
5
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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