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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

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

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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For the part circled in pink, where did 1/4 come from?

73. k=016k² +8 k-3 M8 1.
Transcribed Image Text:73. k=016k² +8 k-3 M8 1.
24 (a ) 10.3.73 16k2+8k-3 = (4k+3)(4k-1), so 1 1 1 1 Thus 16k2 + 8k – 3 (4k + 3)(4k – 1) 4. 4k –1 4k +3 - 1 1 1 -1- the series given is equal to 4. k=0 SO This series telescopes, so Sn 4n - 1 4k –1 4k +3 1 the sum of the series is equal to lim Sn %3D 4 st 1.
Transcribed Image Text:24 (a ) 10.3.73 16k2+8k-3 = (4k+3)(4k-1), so 1 1 1 1 Thus 16k2 + 8k – 3 (4k + 3)(4k – 1) 4. 4k –1 4k +3 - 1 1 1 -1- the series given is equal to 4. k=0 SO This series telescopes, so Sn 4n - 1 4k –1 4k +3 1 the sum of the series is equal to lim Sn %3D 4 st 1.
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