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
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Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 41E
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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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