Part B Here are three attempted inductive proofs of the correct theorem. None of them are correct. For each one, explain what's wrong. "Proof 1:" Suppose that n = 1. Then, k+1 n i=0 2-2 i = i=0 n i=1 = 1 i as was to be shown. This completes the proof. "Proof 2:" Suppose that the theorem holds for some n = k. We'll show that this implies that it also holds for n=k + 1. We can calculate + (k+ 1) 1(1+1) 2 k Σi=i+(k+1) i=0 k(k+ 1) 2 (k+ 1) (k+2) 2 (k+ 1)((k+ 1) + 1) 2 This shows that the statement is true for n=k + 1, which completes the proof. 2 (manipulating sum) (inductive hypothesis) (algebra) (algebra).
Part B Here are three attempted inductive proofs of the correct theorem. None of them are correct. For each one, explain what's wrong. "Proof 1:" Suppose that n = 1. Then, k+1 n i=0 2-2 i = i=0 n i=1 = 1 i as was to be shown. This completes the proof. "Proof 2:" Suppose that the theorem holds for some n = k. We'll show that this implies that it also holds for n=k + 1. We can calculate + (k+ 1) 1(1+1) 2 k Σi=i+(k+1) i=0 k(k+ 1) 2 (k+ 1) (k+2) 2 (k+ 1)((k+ 1) + 1) 2 This shows that the statement is true for n=k + 1, which completes the proof. 2 (manipulating sum) (inductive hypothesis) (algebra) (algebra).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Can you please help me solve this induction problem? Thank you
Transcribed Image Text:Part B
Here are three attempted inductive proofs of the correct theorem. None of them are correct. For each one, explain what's wrong.
"Proof 1:" Suppose that n =
= 1. Then,
k+1
n
i=0
i=0
k
i = Σi + (k + 1)
i=0
k(k+ 1)
2 =
=
2
(k+ 1)(k + 2)
2
=
n
1
as was to be shown. This completes the proof.
"Proof 2:" Suppose that the theorem holds for some n = k. We'll show that this implies that it also holds for n = k + 1. We can
calculate
+ (k+1)
i
1(1+1)
2
(k+ 1)((k+1)+1)
2
9
(manipulating sum)
(inductive hypothesis)
(algebra)
(algebra).
This shows that the statement is true for n = k + 1, which completes the proof.
Transcribed Image Text:"Proof 3:" First, we'll do the base case. Suppose that n =
So, the statement is true for n
=
k+1
Σi
i=0
=
=
1. Then,
n
i=0
1, completing the base case. Now we'll do the inductive step. Suppose that the statement is true for
n=k+1. We'll show that it's also true for n k. We can calculate
=
+
(k+ 1)(k+2)
2
k(k+ 1)
2
2 =
=
=
n
k=0
n
Σi
i=1
1
2(k+ 1)
2
1(1+1)
2
Problem 2
Use mathematical induction to prove that, for any integer n >= 1,
Since what's left includes the formula for the case n = k, this completes the inductive step.
2k
(inductive hypothesis)
=
(algebra).
2n+1 -1
Expert Solution
Step 1: Statement of Principle of Mathematical induction
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