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A (R - 5) Given the power series E (n - 3) A, xla - 5) n = 2 Use the technique that we covered in this unit to Shift the Index of the Power Series so that the exponent of the power series is simplified. A E (n - 3) A xla - 5) n = 2 x(a - 5) (R + 5) A (r + 5) ** becomes R= -3 E (n- 3) A, xla - 5) becomes E (R + 2) A (r + 5) ** R = -3 n = 2 © E (n - 3) A x" - 5) becomes £ (R + 8) A (R - 5) n = 2 R = 7 OE (n- 3) A, xl* - 5) becomes E (R - 8) A (r – 5) ** n = 2 R= 0 00 E (n- 3) A, xla - 5) becomes E (R+ 2) A (R + 5) ** n = 2 R= 2 E (n- 3) A x - 5) becomes (R - 3) A, xe

A (R - 5) Given the power series E (n - 3) A, xla - 5) n = 2 Use the technique that we covered in this unit to Shift the Index of the Power Series so that the exponent of the power series is simplified. A E (n - 3) A xla - 5) n = 2 x(a - 5) (R + 5) A (r + 5) ** becomes R= -3 E (n- 3) A, xla - 5) becomes E (R + 2) A (r + 5) ** R = -3 n = 2 © E (n - 3) A x" - 5) becomes £ (R + 8) A (R - 5) n = 2 R = 7 OE (n- 3) A, xl* - 5) becomes E (R - 8) A (r – 5) ** n = 2 R= 0 00 E (n- 3) A, xla - 5) becomes E (R+ 2) A (R + 5) ** n = 2 R= 2 E (n- 3) A x - 5) becomes (R - 3) A, xe

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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Given the power series 2 (n - 3) A xn – 5) n = 2 Use the technique that we covered in this unit to Shift the Iindex of the Power Series so that the exponent of the power series is simplified. 00 00 (A) 2 (n - 3) A x(" - 5) becomes 2 (R + 5) A (R + 5) ** n = 2 R = -3 B n = 2 2 (n - 3) A x(* - $) becomes E (R + 2) A (R + 5) * R = -3 © E (R + 8) A (a - 5) * 2 (n - 3) A x" - 5) becomes n = 2 R = 7 00 00 O D 2 (n - 3) A x(a – 5) becomes E (R - 8) A (r – 5) ** n = 2 R = 0 00 e 2 (n - 3) A x" - 5) becomes E (R + 2) A (R + 5) E n = 2 R = 2 00 F2 (n - 3) A x" - 5) becomes Σ (R-3 )A, x' n = 2 R = 2
Transcribed Image Text:Given the power series 2 (n - 3) A xn – 5) n = 2 Use the technique that we covered in this unit to Shift the Iindex of the Power Series so that the exponent of the power series is simplified. 00 00 (A) 2 (n - 3) A x(" - 5) becomes 2 (R + 5) A (R + 5) ** n = 2 R = -3 B n = 2 2 (n - 3) A x(* - $) becomes E (R + 2) A (R + 5) * R = -3 © E (R + 8) A (a - 5) * 2 (n - 3) A x" - 5) becomes n = 2 R = 7 00 00 O D 2 (n - 3) A x(a – 5) becomes E (R - 8) A (r – 5) ** n = 2 R = 0 00 e 2 (n - 3) A x" - 5) becomes E (R + 2) A (R + 5) E n = 2 R = 2 00 F2 (n - 3) A x" - 5) becomes Σ (R-3 )A, x' n = 2 R = 2
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