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Given the power series En (n - 1) A, xla + 2) n =1 Use the technique that we covered in this lesson to Shift the Index of the Power Series so that the exponent of the power series is simplified. A 2 n (n - 1) A xm + 2) becomes (R + 2)² A (R + 2) * n =1 n =1 (n - 1) A xla + 2) becomes R2 A, n =1 n =0 En (n - 1) A, xlu + 2) becomes E n =1 n = 3 O 2n (n - 1) A xa + 2) becomes E (R2 + 2R) A n =1 n = 0 En (n - 1) A xlu + 2) becomes (R - 2) (R – 3) A (R - 2) ** n =1 n = 3

Given the power series En (n - 1) A, xla + 2) n =1 Use the technique that we covered in this lesson to Shift the Index of the Power Series so that the exponent of the power series is simplified. A 2 n (n - 1) A xm + 2) becomes (R + 2)² A (R + 2) * n =1 n =1 (n - 1) A xla + 2) becomes R2 A, n =1 n =0 En (n - 1) A, xlu + 2) becomes E n =1 n = 3 O 2n (n - 1) A xa + 2) becomes E (R2 + 2R) A n =1 n = 0 En (n - 1) A xlu + 2) becomes (R - 2) (R – 3) A (R - 2) ** n =1 n = 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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Given the power series 2 n (n - 1) A x (n + 2) n =1 Use the technique that we covered in this lesson to Shift the Index of the Power Series so that the exponent of the power series is simplified. 00 A) Σ E n (n - 1) A x" + 2) becomes 2 (R + 2)? A (r + 2) ** n =1 n = 1 B R2 A, xk 2 n (n - 1) A x" + 2) becomes R n =1 n = 0 00 2 n (n - 1) A x" + 2) becomes n =1 n = 3 00 00 E n (n - 1) A x + 2) becomes E (R2 + 2R) A, x* n =1 n = 0 > n (n - 1) A x" + 2) becomes E (R - 2) (R - 3) A (R – 2) E n =1 n = 3 00 F) Σ 2 n (n - 1) A x" + 2) becomes 2 (R + 2)2 A (R + 2) Σ (R+ 2)? n =1 n = 3
Transcribed Image Text:Given the power series 2 n (n - 1) A x (n + 2) n =1 Use the technique that we covered in this lesson to Shift the Index of the Power Series so that the exponent of the power series is simplified. 00 A) Σ E n (n - 1) A x" + 2) becomes 2 (R + 2)? A (r + 2) ** n =1 n = 1 B R2 A, xk 2 n (n - 1) A x" + 2) becomes R n =1 n = 0 00 2 n (n - 1) A x" + 2) becomes n =1 n = 3 00 00 E n (n - 1) A x + 2) becomes E (R2 + 2R) A, x* n =1 n = 0 > n (n - 1) A x" + 2) becomes E (R - 2) (R - 3) A (R – 2) E n =1 n = 3 00 F) Σ 2 n (n - 1) A x" + 2) becomes 2 (R + 2)2 A (R + 2) Σ (R+ 2)? n =1 n = 3
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