Add this term to both sides of the equation. 20 + 40 +60++20k = 10k(k + 1) 20 + 40 +60+...+20k + (k+ 1) = 10k(k+ 1) +20k Factor the term (k+ 1) out of each term on the right. Note that it is not necessary to write 20k on the left. 20 + 40 +60+.+20k+20(k+ 1) = 10k(k+ 1) +20(k+ 1) 20 + 40 +60+...+20(k+ 1) = (k+ 1)() Now simplify the right side of the equation by factoring out the common factor. 20 + 40 +60+...+20(k+ 1) = (k+ 1)(10k +20) 20+40+ 60+...+20(k+ 1) = (k+1) This final statement is Sk+ 1. Have we finished the proof? O No, as we must show Sn No, as we must show Sn + 1. No, as we have only shown that Sk +1 is true. Yes.
Add this term to both sides of the equation. 20 + 40 +60++20k = 10k(k + 1) 20 + 40 +60+...+20k + (k+ 1) = 10k(k+ 1) +20k Factor the term (k+ 1) out of each term on the right. Note that it is not necessary to write 20k on the left. 20 + 40 +60+.+20k+20(k+ 1) = 10k(k+ 1) +20(k+ 1) 20 + 40 +60+...+20(k+ 1) = (k+ 1)() Now simplify the right side of the equation by factoring out the common factor. 20 + 40 +60+...+20(k+ 1) = (k+ 1)(10k +20) 20+40+ 60+...+20(k+ 1) = (k+1) This final statement is Sk+ 1. Have we finished the proof? O No, as we must show Sn No, as we must show Sn + 1. No, as we have only shown that Sk +1 is true. Yes.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Add this term to both sides of the equation.
.... +20k = 10k(k+ 1)
20 + 40 +60+ ... +20k + (k+ 1) = 10k(k+ 1) +20k
20 + 40 +60+.
Factor the term (k+ 1) out of each term on the right. Note that it is not necessary to write 20k on the left.
20 + 40 +60 + ... +20k + 20(k+ 1) =
10k(k+ 1) +20(k + 1)
1)(0)
20 + 40 +60+...+20(k+ 1) = (k+1)
Now simplify the right side of the equation by factoring out the common factor.
20 + 40 +60 + ... +20(k+ 1) = (k+
20 + 40 +60 + + 20(k+ - 1)
(k+1)
+
This final statement is Sk + 1. Have we finished the proof?
No, as we must show Sn.
O
No, as we must show Sn + 1.
O No, as we have only shown that Sk+1 is true.
Yes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd341268f-2cc2-4ad7-9a20-19238d1cae49%2Fc3b67fdc-cfb2-409b-8f0f-e505a0798214%2Fcxahmpl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Add this term to both sides of the equation.
.... +20k = 10k(k+ 1)
20 + 40 +60+ ... +20k + (k+ 1) = 10k(k+ 1) +20k
20 + 40 +60+.
Factor the term (k+ 1) out of each term on the right. Note that it is not necessary to write 20k on the left.
20 + 40 +60 + ... +20k + 20(k+ 1) =
10k(k+ 1) +20(k + 1)
1)(0)
20 + 40 +60+...+20(k+ 1) = (k+1)
Now simplify the right side of the equation by factoring out the common factor.
20 + 40 +60 + ... +20(k+ 1) = (k+
20 + 40 +60 + + 20(k+ - 1)
(k+1)
+
This final statement is Sk + 1. Have we finished the proof?
No, as we must show Sn.
O
No, as we must show Sn + 1.
O No, as we have only shown that Sk+1 is true.
Yes.
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