The inductive step of an inductive proof shows that for k > 0, if Στo 2 = 2k+1 – 1, then Σ± 20 = 2k+2 – 1. In which step of the proof is the inductive hypothesis used? Step 4 Ο Step 1 Ο Step 3 Step 2 k+1 Στις 23 Σ* +1 23 Στο 23 k+1 = Σ+ 2 = 2-2 – 1 2³ = ₁02³ +2k+1 = (2k+1 -1) + 2k+1 = 2.2k+1 – 1 D (Step 1) (Step 2) (Step 3) (Step 4)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.4: Mathematical Induction
Problem 30E
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The inductive step of an inductive proof shows that for k > 0, if
E-0 23 = 2k+1 -1, then 2 = 2k+2
1. In which step of
j=0
the proof is the inductive hypothesis used?
Step 4
Step 1
Step 3
Step 2
Σ+ 2 = Στο 2 + 2h+1
j=0
j=0
Σ+1 20
j=0
k+1
j=0
Samman
(2k+1 − 1) + 2k+1
=
k+1 2 2.2+1 -1
j=0
Σ+ 2 = 2-2 – 1
j=0
D
(Step 1)
(Step 2)
(Step 3)
(Step 4)
Transcribed Image Text:The inductive step of an inductive proof shows that for k > 0, if E-0 23 = 2k+1 -1, then 2 = 2k+2 1. In which step of j=0 the proof is the inductive hypothesis used? Step 4 Step 1 Step 3 Step 2 Σ+ 2 = Στο 2 + 2h+1 j=0 j=0 Σ+1 20 j=0 k+1 j=0 Samman (2k+1 − 1) + 2k+1 = k+1 2 2.2+1 -1 j=0 Σ+ 2 = 2-2 – 1 j=0 D (Step 1) (Step 2) (Step 3) (Step 4)
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