Tutorial Exercise Find the sum. 4(k + 1)²(k – 4) k = 0 Step 1 Recall the definition of summation notation. az + az + a4 + ... k = 1 For the given problem, k begins with 0 and ends with 3. Therefore, 4(k + 1)2(k – 4) is the sum of 4 4 terms in this series, k = 0 Step 2 Write the terms of the summation. E 4(k + 1)2(k – 4) = {4c0 + 1)²co – - 4)} + {«(1 + 1)²([ ]× - 4)} + {4(2 + 1)?(2- - 4)} + {«c[ ]× +1)² ([ ]x - 4)} k = 0
Tutorial Exercise Find the sum. 4(k + 1)²(k – 4) k = 0 Step 1 Recall the definition of summation notation. az + az + a4 + ... k = 1 For the given problem, k begins with 0 and ends with 3. Therefore, 4(k + 1)2(k – 4) is the sum of 4 4 terms in this series, k = 0 Step 2 Write the terms of the summation. E 4(k + 1)2(k – 4) = {4c0 + 1)²co – - 4)} + {«(1 + 1)²([ ]× - 4)} + {4(2 + 1)?(2- - 4)} + {«c[ ]× +1)² ([ ]x - 4)} k = 0
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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