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5 k3 d. + 2225 k k=1 k=1 п е. + 2n \n k=1 3.

5 k3 d. + 2225 k k=1 k=1 п е. + 2n \n k=1 3.

Calculus: Early Transcendentals
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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Evaluate the sums. a. 2(3 – k?) k=1 6. b. > (k? – 5) - k=1 5 > k(3k + 5) С. k=1 3 5 k3 + 225 5 d. k k=1 \k=1 n EG + 2n) е. k=1
Transcribed Image Text:Evaluate the sums. a. 2(3 – k?) k=1 6. b. > (k? – 5) - k=1 5 > k(3k + 5) С. k=1 3 5 k3 + 225 5 d. k k=1 \k=1 n EG + 2n) е. k=1
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