To continue the proof by induction, begin with 4 + 16 +64 ++4 OA. 4-1 0 в. 4к OC. 4k+1 OD. 4K-1 Add this term to both sides of the equation. 4k+1-4 3 4k+1 3 4+16+64 ++4k = 4+ 16 +64 +...+4K + -4 3 which is assumed to be true, and add the k+ 1 term to both sides of the formula. What is this term? Simplify the right side so that it looks similar to the right side of the equation in the original statement. 4+16+64 +...+4K +4k+ 1 = (Do not factor.) The right side of the above equation is Sk+1. Is the proof complete? OA. No, since it must be shown that Sn+ 1 satisfies the statement. OB. No, since it must be shown that S, satisfies the statement. OC. Yes. OD. No, since it has only been shown that Sk+ 1 satisfies the statement.
To continue the proof by induction, begin with 4 + 16 +64 ++4 OA. 4-1 0 в. 4к OC. 4k+1 OD. 4K-1 Add this term to both sides of the equation. 4k+1-4 3 4k+1 3 4+16+64 ++4k = 4+ 16 +64 +...+4K + -4 3 which is assumed to be true, and add the k+ 1 term to both sides of the formula. What is this term? Simplify the right side so that it looks similar to the right side of the equation in the original statement. 4+16+64 +...+4K +4k+ 1 = (Do not factor.) The right side of the above equation is Sk+1. Is the proof complete? OA. No, since it must be shown that Sn+ 1 satisfies the statement. OB. No, since it must be shown that S, satisfies the statement. OC. Yes. OD. No, since it has only been shown that Sk+ 1 satisfies the statement.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 39E
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