choose the correct answers(few): 1) The proof is correct. 2) The proof is wrong. 3)The predicate is not valid. 4)The base case is wrong. 5)The Inductive Hypothesis is written incorrectly. 6)There is a problem in the Inductive Step.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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choose the correct answers(few):

1) The proof is correct.

2) The proof is wrong.

3)The predicate is not valid.

4)The base case is wrong.

5)The Inductive Hypothesis is written incorrectly.

6)There is a problem in the Inductive Step.

Proof: Let P(n): the sum of the first n powers of two is 2"– 1.
We will prove by induction that P(n) holds for all n e N.
Base Case: P(0) = 2º – 1 = 0.
%3D
|
Since the sum of zero numbers is 0, P(0) holds.
IH: Suppose that P(k) holds, i.e. that
2° + 2'+...+2-1 – 2k – 1
IS: We will show that P(k+1) also holds, meaning
the sum of the first k+1 powers of 2 is 2*+1 – 1
We proceed as follows:
2° + 2'+...+2k-1 + 2k = 2k+1 – 1
. +2*-1
2° + 2'+...+2k–1
2º + 2'+...+2*-1 + 2k = 2* + 2k – 1
2° + 2'+...+2k-1 – 2k – 1
+ 2* = 2(2*) – 1
We have arrived at the IH, which we know to be true.
Therefore P(k+1) holds, completing the induction
Transcribed Image Text:Proof: Let P(n): the sum of the first n powers of two is 2"– 1. We will prove by induction that P(n) holds for all n e N. Base Case: P(0) = 2º – 1 = 0. %3D | Since the sum of zero numbers is 0, P(0) holds. IH: Suppose that P(k) holds, i.e. that 2° + 2'+...+2-1 – 2k – 1 IS: We will show that P(k+1) also holds, meaning the sum of the first k+1 powers of 2 is 2*+1 – 1 We proceed as follows: 2° + 2'+...+2k-1 + 2k = 2k+1 – 1 . +2*-1 2° + 2'+...+2k–1 2º + 2'+...+2*-1 + 2k = 2* + 2k – 1 2° + 2'+...+2k-1 – 2k – 1 + 2* = 2(2*) – 1 We have arrived at the IH, which we know to be true. Therefore P(k+1) holds, completing the induction
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