q8 Given the power series2 2n A x(m + 3)n = 0Use the technique that we covered in this unit to Shift the Index of the Power Series so that the exponent of the power series issimplified.00A> 2n A x" + 3) becomes 2 2 R A, x"n = 0R = 0B 2 2n AE 2n A x" + 3) becomes 2 2 (R + 3) A (R + 3)(n + 3)n = 0R = 0002 2n A x(n + 3)2 2 (R + 3) A (R + 3) **becomes >n = 0R = -300Σ Α,2 2n A x(m(n + 3) becomes E 2 (R -3) A (R - 3) **n = 0R = 0E> 2n A x" + 3) becomes 2 (R - 3) A (R - 3)n = 0R = 32n A x" + 3) becomes 2 2(R + 3) A, x*n = 0R = 0

To shift the index of the power series use the substitution n = R - 3

Given the power series 2 2n A x(n + 3) n = 0 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 > x" 2 2n A x" + 3) becomes 2 2 R A, n = 0 R = 0 B E 2n A x" + 3) becomes 2 2 (R + 3) A (R + 3) * n = 0 R = 0 00 © E 2n A, xl" + 3) becomes 2 2 (R+ 3) A (R + 3) *"* n = 0 R = -3 00 ΟΣ E 2n A x(m + 3) becomes 2 (R - 3) A (R - 3) n = 0 R = 0 > 2n A x(n + 3) becomes 2 (R - 3) A R - 3) n =0 R = 3 00 2n A x(" + 3) becomes 2 (R + 3) A, xR n =0 R = 0

Given the power series 2 2n A x(n + 3) n = 0 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 > x" 2 2n A x" + 3) becomes 2 2 R A, n = 0 R = 0 B E 2n A x" + 3) becomes 2 2 (R + 3) A (R + 3) * n = 0 R = 0 00 © E 2n A, xl" + 3) becomes 2 2 (R+ 3) A (R + 3) " n = 0 R = -3 00 ΟΣ E 2n A x(m + 3) becomes 2 (R - 3) A (R - 3) n = 0 R = 0 > 2n A x(n + 3) becomes 2 (R - 3) A R - 3) n =0 R = 3 00 2n A x(" + 3) becomes 2 (R + 3) A, xR n =0 R = 0

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.2: Arithmetic Sequences

Problem 68E

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