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
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 41E
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Question
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
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