For the part circled in pink, where did 1/4 come from? 73.k=016k² +8 k-3M81. 24(a )10.3.73 16k2+8k-3 = (4k+3)(4k-1), so1111Thus16k2 + 8k – 3 (4k + 3)(4k – 1)4.4k –1 4k +3-111-1-the series given is equal to4.k=0SOThis series telescopes, so Sn4n - 14k –1 4k +31the sum of the series is equal to lim Sn%3D4st1.

Answered: Image /qna-images/answer/9db2c4d4-1444-45d6-81c2-73771bad7e23.jpg

Answered: 1 1 1 Thus 10.3.73 16k2+8k-3 =…

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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1 1 1 Thus 10.3.73 16k2+8k-3 = (4k+3)(4k-1), so 4k – 1 4k +3 4 16k2 + 8k - 3 (4k +3)(4k - 1) 1 -1- 1 1. SO %3D This series telescopes, so Sn 4n – 1 the series given is equal to 4. k=0 4k - 1 4k +3 1 the sum of the series is equal to lim Sn = 4 n-00