\ 1. 2 + 4 + 6 + · · · + 2n = n(n + 1) 2. 1 + 5 + 9 + · ·· + (4n – 3) = n(2n – 1) 3. 3 + 4 + 5 + .. + (n + 2) = ;n(n + 5) п (п + 5) 4. 3 + 5 + 7 + ... + (2n + 1) = n(n + 2) 1 5. 2 + 5 + 8 + · + (3n – 1) =;n(3n + 1) 6. 1+ 4 + 7 + · + (3n – 2) = n(3n – 1)
\ 1. 2 + 4 + 6 + · · · + 2n = n(n + 1) 2. 1 + 5 + 9 + · ·· + (4n – 3) = n(2n – 1) 3. 3 + 4 + 5 + .. + (n + 2) = ;n(n + 5) п (п + 5) 4. 3 + 5 + 7 + ... + (2n + 1) = n(n + 2) 1 5. 2 + 5 + 8 + · + (3n – 1) =;n(3n + 1) 6. 1+ 4 + 7 + · + (3n – 2) = n(3n – 1)
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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use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers n.
![\ 1. 2 + 4 + 6 + · · · + 2n = n(n + 1)
2. 1 + 5 + 9 + · ·· + (4n – 3) = n(2n – 1)
3. 3 + 4 + 5 + .. + (n + 2) = ;n(n + 5)
п (п + 5)
4. 3 + 5 + 7 +
... + (2n + 1) = n(n + 2)
1
5. 2 + 5 + 8 + · + (3n – 1) =;n(3n + 1)
6. 1+ 4 + 7 + · + (3n – 2) = n(3n – 1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fafa6fbd1-79d8-41da-8d76-0f495e9ac1a1%2F3cf8363b-caf6-4fc0-99ce-23ec67f93e31%2Fwz1nw0d_processed.png&w=3840&q=75)
Transcribed Image Text:\ 1. 2 + 4 + 6 + · · · + 2n = n(n + 1)
2. 1 + 5 + 9 + · ·· + (4n – 3) = n(2n – 1)
3. 3 + 4 + 5 + .. + (n + 2) = ;n(n + 5)
п (п + 5)
4. 3 + 5 + 7 +
... + (2n + 1) = n(n + 2)
1
5. 2 + 5 + 8 + · + (3n – 1) =;n(3n + 1)
6. 1+ 4 + 7 + · + (3n – 2) = n(3n – 1)
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