Prove the statement by mathematical induction. 3+5+7+…+(2n+1)=n(n+2) 1. Proposition is true when n=1, since n(n+2)=A(1+2)=B. A= B= 2. We will assume that the proposition is true for a constant k=n, so 3+5+7+…+(2k+1)=C(k+D). C= D= 3. Then, 3+5+7+…+(2k+1)+(Ek+F)=k(k+2)+(Gk+H). E= F= G= H= 4. Thus: ∑i=1k+12k+1=k2+4k+3

Prove the statement by mathematical induction. 3+5+7+…+(2n+1)=n(n+2) 1. Proposition is true when n=1, since n(n+2)=A(1+2)=B. A= B= 2. We will assume that the proposition is true for a constant k=n, so 3+5+7+…+(2k+1)=C(k+D). C= D= 3. Then, 3+5+7+…+(2k+1)+(Ek+F)=k(k+2)+(Gk+H). E= F= G= H= 4. Thus: ∑i=1k+12k+1=k2+4k+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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