Read the following proof and identify proof steps. 1. Here we will prove that the sum of the first n odd integers is n2, That is, Vn > 0 E,(2n – 1) = n? 'i=1 2. We will use mathematical induction on Ση 1(2n – 1) = n? | 3. For n=1, EL (2 · 1 – 1) = 2 · 1 'i=1 Base case -1=1= 12 Induction step 4. Assume that E (2k – 1) = k² Theorem statement 5. E (2(k+1) – 1) = (E (2k – 1)) + (2(k+1) – 1) = k² + (2(k+1) – 1) = k² + 2k +1 = (k + 1)² Induction hypothesis 3=1 %3D

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 63RE
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Read the following proof and identify proof steps.
1. Here we will prove that the sum of
the first n odd integers is n2, That
is, Vn > 0 E,(2n – 1)
= n?
i=1
2. We will use mathematical
induction on
E (2n – 1) = n²
i=1
3. For n=1,
E (2 ·1– 1) = 2 · 1
-1=1= 12
'i=1
Base case
Induction step
4. Assume that
E(2k – 1) = k²
Theorem statement
5.
E" (2(k+1) – 1)
= (E(2k – 1))
+ (2(k+1) – 1) = k²
+ (2(k+1) – 1) = k²
+ 2k +1= (k+ 1)²
Induction hypothesis
i31
6. Therefore,
Vn > 0 E,(2n – 1)
= n?
Transcribed Image Text:Read the following proof and identify proof steps. 1. Here we will prove that the sum of the first n odd integers is n2, That is, Vn > 0 E,(2n – 1) = n? i=1 2. We will use mathematical induction on E (2n – 1) = n² i=1 3. For n=1, E (2 ·1– 1) = 2 · 1 -1=1= 12 'i=1 Base case Induction step 4. Assume that E(2k – 1) = k² Theorem statement 5. E" (2(k+1) – 1) = (E(2k – 1)) + (2(k+1) – 1) = k² + (2(k+1) – 1) = k² + 2k +1= (k+ 1)² Induction hypothesis i31 6. Therefore, Vn > 0 E,(2n – 1) = n?
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