SlideShare a Scribd company logo
1 of 45
Mathematical Induction
Lecture 11, CMSC 56
Allyn Joy D. Calcaben
Find the sum and the product of each pairs of numbers. Express
your answers as a binary expansion.
1. (100 0111)2 , (111 0111)2
2. (1110 1111)2 , (1011 1101)2
3. (10 1010 1010)2 , (1 1111 0000)2
Assignment (1 whole Sheet Paper)
Suppose that we have an infinite ladder, as
shown in the figure, and we want to know
whether we can reach every step on this ladder.
We know two things:
1. We can reach the first rung of the ladder.
2. If we can reach a particular rung of the
ladder, then we can reach the next rung.
Can we conclude that we are able to
reach every rung of this infinite
ladder?
The answer is yes, something we can verify using an important
proof technique called mathematical induction.
To prove that P(n) is true for all positive integers n, where P(n) is a
propositional function, we complete two steps:
Principle of Mathematical Induction
To prove that P(n) is true for all positive integers n, where P(n) is a
propositional function, we complete two steps:
Basis Step: We verify that P(1) is true.
Inductive Step: We show that the conditional statement
P(k) → P(k + 1) is true for all positive integers k.
Principle of Mathematical Induction
The assumption that P(k) is true is called the inductive hypothesis.
When we use mathematical induction to prove a theorem, we first
show that P(1) is true. Then we know that P(2) is true, because
P(1) → P(2). Further, we know that P(3) is true, P(2) → P(3).
Continuing along these lines, we see that P(n) is true for every
positive integer n.
Principle of Mathematical Induction
1. We can reach the first rung of the ladder.
2. If we can reach a particular rung of the ladder, then we reach
the next rung.
Infinite Ladder
1. We can reach the first rung of the ladder.
2. If we can reach a particular rung of the ladder, then we reach
the next rung.
Basis Step: Verify that P(1) is true.
Inductive Step: P(k) → P(k + 1) is true for all positive integers k
Therefore, P(n) is true for all positive integers n, where P(n) is the statement
that we can reach the nth rung of the ladder. Consequently, we can invoke
mathematical induction to conclude that we can reach every rung.
Infinite Ladder
How Mathematical Induction Works
“Domino Effect”
Step 1. The first domino falls
Step 2. When any domino falls, the next domino falls
“Domino Effect”
Step 1. The first domino falls
Step 2. When any domino falls, the next domino falls
….. all dominos will fall!
This is how Mathematical Induction works.
Example
There are infinitely many stations on a train route. Suppose
that the train stops at the first station and suppose that if the train
stops at a station, then is stops at the next station. Show that the
train stops at all stations.
Solution
Let P(n) be the statement that the train stops at station n.
Basis Step: We are told that P(1) is true.
Inductive Step: We are told that P(k) implies P(k+1) for each n ≥ 1.
Therefore by the principle of the mathematical induction, P(n) is
true for all positive integers n.
Proving Summation Formulae
Mathematical Induction is particularly well suited for proving that
such formulae are valid. However, summation formulae can be
proven in other ways.
major disadvantage of using mathematical induction:
1. you cannot use it to derive this formula.
Example
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Let P (n) be the proposition that the sum of the first n positive integers, 1 + 2
+ ··· n =
𝑛 (𝑛+1)
2
, is
𝑛 (𝑛+1)
2
. We must do two things to prove that P (n) is true
for n = 1, 2, 3,.... Namely, we must show that P(1) is true and that the
conditional statement P(k)implies P(k+1) is true for k = 1, 2, 3,....
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: We show that the conditional statement
P(k) → P(k + 1) is true for all positive integers k.
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: First, assume P(k) is true for all positive integers k.
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: First, assume P(k) is true for all positive integers k.
(Inductive Hypothesis)
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Inductive Hypothesis:
1 + 2 + . . . + 𝑘 =
𝑘 (𝑘+1)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Under this assumption, it must be shown that
P(𝑘 + 1) is true.
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Under this assumption, it must be shown that
P(𝑘 + 1) is true.
1 + 2 + . . . + 𝑘 + (𝑘 + 1) =
𝑘+1 [(𝑘+1)+1]
2
=
𝑘+1 (𝑘+2)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘)
P(𝑘): 1 + 2 + . . . + 𝑘 =
𝑘 (𝑘+1)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘)
Left Side:
P(𝑘): 1 + 2 + . . . + 𝑘 =
𝑘 (𝑘+1)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘)
Left Side:
P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) =
𝑘 (𝑘+1)
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘)
Right Side:
P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) =
𝑘 (𝑘+1)
2
+ (𝑘 + 1)
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘)
P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) =
𝑘 𝑘+1 + 2 𝑘+1
2
Solution
Show that if 𝑛 is a positive integer, then
1 + 2 + . . . + 𝑛 =
𝑛 (𝑛+1)
2
Basis Step: P(1) is true, because 1 =
1 (1+1)
2
Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘)
P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) =
𝑘+1 𝑘+2
2
Example
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Let P(𝑛) be the proposition that 1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for the integer 𝑛.
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Assume P(k) is true for an arbitrary nonnegative integer k.
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Assume P(k) is true for an arbitrary nonnegative integer k.
(Inductive Hypothesis)
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Inductive Hypothesis:
1 + 2 + 22 + . . . + 2 𝑘 = 2
(𝑘 + 1)
− 1
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is
true.
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is
true.
1 + 2 + 22 + . . . + 2 𝑘 + 2
(𝑘 + 1)
= 2
(𝑘 + 1) + 1
−1 = 2
(𝑘 + 2)
−1
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Add 2
(𝑘 + 1)
to both sides of the equation in P(𝑘)
P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 = 2
(𝑘 + 1)
− 1
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Add 2
(𝑘 + 1)
to both sides of the equation in P(𝑘)
Left Side:
P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2
(𝑘 + 1)
= 2
(𝑘 + 1)
− 1
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Add 2
(𝑘 + 1)
to both sides of the equation in P(𝑘)
Right Side:
P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2
(𝑘 + 1)
= 2
(𝑘 + 1)
− 1 + 2
(𝑘 + 1)
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Add 2
(𝑘 + 1)
to both sides of the equation in P(𝑘)
P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2
(𝑘 + 1)
= 2 ⋅ 2
(𝑘 + 1)
− 1
Solution
Use mathematical induction to show that
1 + 2 + 22 + . . . + 2 𝑛 = 2
(𝑛 + 1)
− 1
for all nonnegative integers 𝑛.
Basis Step: Let P(0) is true, because 20 = 1 = 2
1
− 1
Inductive Step: Add 2
(𝑘 + 1)
to both sides of the equation in P(𝑘)
P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2
(𝑘 + 1)
= 2
(𝑘 + 2)
− 1
1. Prove that 12 + 32 + 52 + . . . + (2𝑛 + 1) 2 =
( 𝑛 + 1) (2 𝑛 + 1)(2 𝑛 + 3)
3
whenever 𝑛 is a nonnegative integer.
2. Prove that 3 + 3 ⋅ 5 + 3 ⋅ 52 + . . . + 3 ⋅ 5n =
3(5
𝑛 +1
− 1)
4
whenever
𝑛 is a nonnegative integer.
Challenge (1/2 Sheet Paper)

More Related Content

What's hot

CMSC 56 | Lecture 5: Proofs Methods and Strategy
CMSC 56 | Lecture 5: Proofs Methods and StrategyCMSC 56 | Lecture 5: Proofs Methods and Strategy
CMSC 56 | Lecture 5: Proofs Methods and Strategyallyn joy calcaben
 
Inner product spaces
Inner product spacesInner product spaces
Inner product spacesEasyStudy3
 
Rules of inference
Rules of inferenceRules of inference
Rules of inferenceLakshmi R
 
21 monotone sequences x
21 monotone sequences x21 monotone sequences x
21 monotone sequences xmath266
 
5 4 function notation
5 4 function notation5 4 function notation
5 4 function notationhisema01
 
Infinite series
Infinite seriesInfinite series
Infinite seriesjaflint718
 
CMSC 56 | Lecture 1: Propositional Logic
CMSC 56 | Lecture 1: Propositional LogicCMSC 56 | Lecture 1: Propositional Logic
CMSC 56 | Lecture 1: Propositional Logicallyn joy calcaben
 
Binomial Theorem
Binomial TheoremBinomial Theorem
Binomial Theoremitutor
 
CMSC 56 | Lecture 4: Rules of Inference
CMSC 56 | Lecture 4: Rules of InferenceCMSC 56 | Lecture 4: Rules of Inference
CMSC 56 | Lecture 4: Rules of Inferenceallyn joy calcaben
 
CMSC 56 | Lecture 6: Sets & Set Operations
CMSC 56 | Lecture 6: Sets & Set OperationsCMSC 56 | Lecture 6: Sets & Set Operations
CMSC 56 | Lecture 6: Sets & Set Operationsallyn joy calcaben
 
34 polar coordinate and equations
34 polar coordinate and equations34 polar coordinate and equations
34 polar coordinate and equationsmath266
 
Inner Product Space
Inner Product SpaceInner Product Space
Inner Product SpacePatel Raj
 
Composition Of Functions
Composition Of FunctionsComposition Of Functions
Composition Of Functionssjwong
 

What's hot (20)

Proof By Contradictions
Proof By ContradictionsProof By Contradictions
Proof By Contradictions
 
CMSC 56 | Lecture 5: Proofs Methods and Strategy
CMSC 56 | Lecture 5: Proofs Methods and StrategyCMSC 56 | Lecture 5: Proofs Methods and Strategy
CMSC 56 | Lecture 5: Proofs Methods and Strategy
 
Inner product spaces
Inner product spacesInner product spaces
Inner product spaces
 
Rules of inference
Rules of inferenceRules of inference
Rules of inference
 
21 monotone sequences x
21 monotone sequences x21 monotone sequences x
21 monotone sequences x
 
5 4 function notation
5 4 function notation5 4 function notation
5 4 function notation
 
Infinite series
Infinite seriesInfinite series
Infinite series
 
CMSC 56 | Lecture 1: Propositional Logic
CMSC 56 | Lecture 1: Propositional LogicCMSC 56 | Lecture 1: Propositional Logic
CMSC 56 | Lecture 1: Propositional Logic
 
Binomial Theorem
Binomial TheoremBinomial Theorem
Binomial Theorem
 
Recurrence relation
Recurrence relationRecurrence relation
Recurrence relation
 
Logic&proof
Logic&proofLogic&proof
Logic&proof
 
CMSC 56 | Lecture 4: Rules of Inference
CMSC 56 | Lecture 4: Rules of InferenceCMSC 56 | Lecture 4: Rules of Inference
CMSC 56 | Lecture 4: Rules of Inference
 
CMSC 56 | Lecture 6: Sets & Set Operations
CMSC 56 | Lecture 6: Sets & Set OperationsCMSC 56 | Lecture 6: Sets & Set Operations
CMSC 56 | Lecture 6: Sets & Set Operations
 
Power series
Power series Power series
Power series
 
Propositional logic
Propositional logicPropositional logic
Propositional logic
 
34 polar coordinate and equations
34 polar coordinate and equations34 polar coordinate and equations
34 polar coordinate and equations
 
Propositional logic
Propositional logicPropositional logic
Propositional logic
 
Inner Product Space
Inner Product SpaceInner Product Space
Inner Product Space
 
Composition Of Functions
Composition Of FunctionsComposition Of Functions
Composition Of Functions
 
Proof by contradiction
Proof by contradictionProof by contradiction
Proof by contradiction
 

Similar to CMSC 56 | Lecture 11: Mathematical Induction

Similar to CMSC 56 | Lecture 11: Mathematical Induction (20)

Mathematical Induction DM
Mathematical Induction DMMathematical Induction DM
Mathematical Induction DM
 
Induction q
Induction qInduction q
Induction q
 
Induction q
Induction qInduction q
Induction q
 
11-Induction CIIT.pptx
11-Induction CIIT.pptx11-Induction CIIT.pptx
11-Induction CIIT.pptx
 
Introduction to mathematical Induction.ppt
Introduction to mathematical Induction.pptIntroduction to mathematical Induction.ppt
Introduction to mathematical Induction.ppt
 
Induction (1).ppt
Induction (1).pptInduction (1).ppt
Induction (1).ppt
 
Chapter 4 dis 2011
Chapter 4 dis 2011Chapter 4 dis 2011
Chapter 4 dis 2011
 
ppt02.ppt
ppt02.pptppt02.ppt
ppt02.ppt
 
mathematical induction and stuff Induction.pptx
mathematical induction and stuff Induction.pptxmathematical induction and stuff Induction.pptx
mathematical induction and stuff Induction.pptx
 
Slide subtopic 5
Slide subtopic 5Slide subtopic 5
Slide subtopic 5
 
Induction.pdf
Induction.pdfInduction.pdf
Induction.pdf
 
mathematicalinductionanddivisibilityrules-160711105713.pdf
mathematicalinductionanddivisibilityrules-160711105713.pdfmathematicalinductionanddivisibilityrules-160711105713.pdf
mathematicalinductionanddivisibilityrules-160711105713.pdf
 
Mathematical induction and divisibility rules
Mathematical induction and divisibility rulesMathematical induction and divisibility rules
Mathematical induction and divisibility rules
 
Chapter 5
Chapter 5Chapter 5
Chapter 5
 
Per4 induction
Per4 inductionPer4 induction
Per4 induction
 
17-mathematical-induction.ppt
17-mathematical-induction.ppt17-mathematical-induction.ppt
17-mathematical-induction.ppt
 
17-mathematical-induction.ppt
17-mathematical-induction.ppt17-mathematical-induction.ppt
17-mathematical-induction.ppt
 
Factorials as sums
Factorials as sumsFactorials as sums
Factorials as sums
 
Olimpiade matematika di kanada 2018
Olimpiade matematika di kanada 2018Olimpiade matematika di kanada 2018
Olimpiade matematika di kanada 2018
 
6e-ch4.ppt
6e-ch4.ppt6e-ch4.ppt
6e-ch4.ppt
 

More from allyn joy calcaben

CMSC 56 | Lecture 17: Matrices
CMSC 56 | Lecture 17: MatricesCMSC 56 | Lecture 17: Matrices
CMSC 56 | Lecture 17: Matricesallyn joy calcaben
 
CMSC 56 | Lecture 16: Equivalence of Relations & Partial Ordering
CMSC 56 | Lecture 16: Equivalence of Relations & Partial OrderingCMSC 56 | Lecture 16: Equivalence of Relations & Partial Ordering
CMSC 56 | Lecture 16: Equivalence of Relations & Partial Orderingallyn joy calcaben
 
CMSC 56 | Lecture 15: Closures of Relations
CMSC 56 | Lecture 15: Closures of RelationsCMSC 56 | Lecture 15: Closures of Relations
CMSC 56 | Lecture 15: Closures of Relationsallyn joy calcaben
 
CMSC 56 | Lecture 14: Representing Relations
CMSC 56 | Lecture 14: Representing RelationsCMSC 56 | Lecture 14: Representing Relations
CMSC 56 | Lecture 14: Representing Relationsallyn joy calcaben
 
CMSC 56 | Lecture 13: Relations and their Properties
CMSC 56 | Lecture 13: Relations and their PropertiesCMSC 56 | Lecture 13: Relations and their Properties
CMSC 56 | Lecture 13: Relations and their Propertiesallyn joy calcaben
 
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program CorrectnessCMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctnessallyn joy calcaben
 
CMSC 56 | Lecture 10: Integer Representations & Algorithms
CMSC 56 | Lecture 10: Integer Representations & AlgorithmsCMSC 56 | Lecture 10: Integer Representations & Algorithms
CMSC 56 | Lecture 10: Integer Representations & Algorithmsallyn joy calcaben
 
CMSC 56 | Lecture 9: Functions Representations
CMSC 56 | Lecture 9: Functions RepresentationsCMSC 56 | Lecture 9: Functions Representations
CMSC 56 | Lecture 9: Functions Representationsallyn joy calcaben
 
CMSC 56 | Lecture 8: Growth of Functions
CMSC 56 | Lecture 8: Growth of FunctionsCMSC 56 | Lecture 8: Growth of Functions
CMSC 56 | Lecture 8: Growth of Functionsallyn joy calcaben
 
CMSC 56 | Lecture 3: Predicates & Quantifiers
CMSC 56 | Lecture 3: Predicates & QuantifiersCMSC 56 | Lecture 3: Predicates & Quantifiers
CMSC 56 | Lecture 3: Predicates & Quantifiersallyn joy calcaben
 
CMSC 56 | Lecture 2: Propositional Equivalences
CMSC 56 | Lecture 2: Propositional EquivalencesCMSC 56 | Lecture 2: Propositional Equivalences
CMSC 56 | Lecture 2: Propositional Equivalencesallyn joy calcaben
 
La Solidaridad and the Propaganda Movement
La Solidaridad and the Propaganda MovementLa Solidaridad and the Propaganda Movement
La Solidaridad and the Propaganda Movementallyn joy calcaben
 
Computer Vision: Feature matching with RANSAC Algorithm
Computer Vision: Feature matching with RANSAC AlgorithmComputer Vision: Feature matching with RANSAC Algorithm
Computer Vision: Feature matching with RANSAC Algorithmallyn joy calcaben
 
Chapter 7 expressions and assignment statements ii
Chapter 7 expressions and assignment statements iiChapter 7 expressions and assignment statements ii
Chapter 7 expressions and assignment statements iiallyn joy calcaben
 

More from allyn joy calcaben (15)

CMSC 56 | Lecture 17: Matrices
CMSC 56 | Lecture 17: MatricesCMSC 56 | Lecture 17: Matrices
CMSC 56 | Lecture 17: Matrices
 
CMSC 56 | Lecture 16: Equivalence of Relations & Partial Ordering
CMSC 56 | Lecture 16: Equivalence of Relations & Partial OrderingCMSC 56 | Lecture 16: Equivalence of Relations & Partial Ordering
CMSC 56 | Lecture 16: Equivalence of Relations & Partial Ordering
 
CMSC 56 | Lecture 15: Closures of Relations
CMSC 56 | Lecture 15: Closures of RelationsCMSC 56 | Lecture 15: Closures of Relations
CMSC 56 | Lecture 15: Closures of Relations
 
CMSC 56 | Lecture 14: Representing Relations
CMSC 56 | Lecture 14: Representing RelationsCMSC 56 | Lecture 14: Representing Relations
CMSC 56 | Lecture 14: Representing Relations
 
CMSC 56 | Lecture 13: Relations and their Properties
CMSC 56 | Lecture 13: Relations and their PropertiesCMSC 56 | Lecture 13: Relations and their Properties
CMSC 56 | Lecture 13: Relations and their Properties
 
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program CorrectnessCMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness
 
CMSC 56 | Lecture 10: Integer Representations & Algorithms
CMSC 56 | Lecture 10: Integer Representations & AlgorithmsCMSC 56 | Lecture 10: Integer Representations & Algorithms
CMSC 56 | Lecture 10: Integer Representations & Algorithms
 
CMSC 56 | Lecture 9: Functions Representations
CMSC 56 | Lecture 9: Functions RepresentationsCMSC 56 | Lecture 9: Functions Representations
CMSC 56 | Lecture 9: Functions Representations
 
CMSC 56 | Lecture 8: Growth of Functions
CMSC 56 | Lecture 8: Growth of FunctionsCMSC 56 | Lecture 8: Growth of Functions
CMSC 56 | Lecture 8: Growth of Functions
 
CMSC 56 | Lecture 3: Predicates & Quantifiers
CMSC 56 | Lecture 3: Predicates & QuantifiersCMSC 56 | Lecture 3: Predicates & Quantifiers
CMSC 56 | Lecture 3: Predicates & Quantifiers
 
CMSC 56 | Lecture 2: Propositional Equivalences
CMSC 56 | Lecture 2: Propositional EquivalencesCMSC 56 | Lecture 2: Propositional Equivalences
CMSC 56 | Lecture 2: Propositional Equivalences
 
La Solidaridad and the Propaganda Movement
La Solidaridad and the Propaganda MovementLa Solidaridad and the Propaganda Movement
La Solidaridad and the Propaganda Movement
 
Computer Vision: Feature matching with RANSAC Algorithm
Computer Vision: Feature matching with RANSAC AlgorithmComputer Vision: Feature matching with RANSAC Algorithm
Computer Vision: Feature matching with RANSAC Algorithm
 
Chapter 7 expressions and assignment statements ii
Chapter 7 expressions and assignment statements iiChapter 7 expressions and assignment statements ii
Chapter 7 expressions and assignment statements ii
 
#11 osteoporosis
#11 osteoporosis#11 osteoporosis
#11 osteoporosis
 

Recently uploaded

Championnat de France de Tennis de table/
Championnat de France de Tennis de table/Championnat de France de Tennis de table/
Championnat de France de Tennis de table/siemaillard
 
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...Denish Jangid
 
2024_Student Session 2_ Set Plan Preparation.pptx
2024_Student Session 2_ Set Plan Preparation.pptx2024_Student Session 2_ Set Plan Preparation.pptx
2024_Student Session 2_ Set Plan Preparation.pptxmansk2
 
Behavioral-sciences-dr-mowadat rana (1).pdf
Behavioral-sciences-dr-mowadat rana (1).pdfBehavioral-sciences-dr-mowadat rana (1).pdf
Behavioral-sciences-dr-mowadat rana (1).pdfaedhbteg
 
Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
 Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatmentsaipooja36
 
會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽
會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽
會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽中 央社
 
philosophy and it's principles based on the life
philosophy and it's principles based on the lifephilosophy and it's principles based on the life
philosophy and it's principles based on the lifeNitinDeodare
 
An Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptxAn Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptxCeline George
 
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdfDanh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdfQucHHunhnh
 
Morse OER Some Benefits and Challenges.pptx
Morse OER Some Benefits and Challenges.pptxMorse OER Some Benefits and Challenges.pptx
Morse OER Some Benefits and Challenges.pptxjmorse8
 
How to Manage Closest Location in Odoo 17 Inventory
How to Manage Closest Location in Odoo 17 InventoryHow to Manage Closest Location in Odoo 17 Inventory
How to Manage Closest Location in Odoo 17 InventoryCeline George
 
MichaelStarkes_UncutGemsProjectSummary.pdf
MichaelStarkes_UncutGemsProjectSummary.pdfMichaelStarkes_UncutGemsProjectSummary.pdf
MichaelStarkes_UncutGemsProjectSummary.pdfmstarkes24
 
會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文
會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文
會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文中 央社
 
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...Nguyen Thanh Tu Collection
 
Capitol Tech Univ Doctoral Presentation -May 2024
Capitol Tech Univ Doctoral Presentation -May 2024Capitol Tech Univ Doctoral Presentation -May 2024
Capitol Tech Univ Doctoral Presentation -May 2024CapitolTechU
 
Discover the Dark Web .pdf InfosecTrain
Discover the Dark Web .pdf  InfosecTrainDiscover the Dark Web .pdf  InfosecTrain
Discover the Dark Web .pdf InfosecTraininfosec train
 
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17Celine George
 
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...Nguyen Thanh Tu Collection
 

Recently uploaded (20)

Word Stress rules esl .pptx
Word Stress rules esl               .pptxWord Stress rules esl               .pptx
Word Stress rules esl .pptx
 
Championnat de France de Tennis de table/
Championnat de France de Tennis de table/Championnat de France de Tennis de table/
Championnat de France de Tennis de table/
 
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
Basic Civil Engineering notes on Transportation Engineering, Modes of Transpo...
 
2024_Student Session 2_ Set Plan Preparation.pptx
2024_Student Session 2_ Set Plan Preparation.pptx2024_Student Session 2_ Set Plan Preparation.pptx
2024_Student Session 2_ Set Plan Preparation.pptx
 
Behavioral-sciences-dr-mowadat rana (1).pdf
Behavioral-sciences-dr-mowadat rana (1).pdfBehavioral-sciences-dr-mowadat rana (1).pdf
Behavioral-sciences-dr-mowadat rana (1).pdf
 
Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
 Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
Envelope of Discrepancy in Orthodontics: Enhancing Precision in Treatment
 
會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽
會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽
會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽會考英聽
 
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdfPost Exam Fun(da) Intra UEM General Quiz - Finals.pdf
Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf
 
philosophy and it's principles based on the life
philosophy and it's principles based on the lifephilosophy and it's principles based on the life
philosophy and it's principles based on the life
 
An Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptxAn Overview of the Odoo 17 Discuss App.pptx
An Overview of the Odoo 17 Discuss App.pptx
 
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdfDanh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
Danh sách HSG Bộ môn cấp trường - Cấp THPT.pdf
 
Morse OER Some Benefits and Challenges.pptx
Morse OER Some Benefits and Challenges.pptxMorse OER Some Benefits and Challenges.pptx
Morse OER Some Benefits and Challenges.pptx
 
How to Manage Closest Location in Odoo 17 Inventory
How to Manage Closest Location in Odoo 17 InventoryHow to Manage Closest Location in Odoo 17 Inventory
How to Manage Closest Location in Odoo 17 Inventory
 
MichaelStarkes_UncutGemsProjectSummary.pdf
MichaelStarkes_UncutGemsProjectSummary.pdfMichaelStarkes_UncutGemsProjectSummary.pdf
MichaelStarkes_UncutGemsProjectSummary.pdf
 
會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文
會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文
會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文會考英文
 
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
BỘ LUYỆN NGHE TIẾNG ANH 8 GLOBAL SUCCESS CẢ NĂM (GỒM 12 UNITS, MỖI UNIT GỒM 3...
 
Capitol Tech Univ Doctoral Presentation -May 2024
Capitol Tech Univ Doctoral Presentation -May 2024Capitol Tech Univ Doctoral Presentation -May 2024
Capitol Tech Univ Doctoral Presentation -May 2024
 
Discover the Dark Web .pdf InfosecTrain
Discover the Dark Web .pdf  InfosecTrainDiscover the Dark Web .pdf  InfosecTrain
Discover the Dark Web .pdf InfosecTrain
 
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
Removal Strategy _ FEFO _ Working with Perishable Products in Odoo 17
 
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
 

CMSC 56 | Lecture 11: Mathematical Induction

  • 1. Mathematical Induction Lecture 11, CMSC 56 Allyn Joy D. Calcaben
  • 2. Find the sum and the product of each pairs of numbers. Express your answers as a binary expansion. 1. (100 0111)2 , (111 0111)2 2. (1110 1111)2 , (1011 1101)2 3. (10 1010 1010)2 , (1 1111 0000)2 Assignment (1 whole Sheet Paper)
  • 3. Suppose that we have an infinite ladder, as shown in the figure, and we want to know whether we can reach every step on this ladder. We know two things: 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we can reach the next rung.
  • 4. Can we conclude that we are able to reach every rung of this infinite ladder?
  • 5. The answer is yes, something we can verify using an important proof technique called mathematical induction.
  • 6. To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps: Principle of Mathematical Induction
  • 7. To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, we complete two steps: Basis Step: We verify that P(1) is true. Inductive Step: We show that the conditional statement P(k) → P(k + 1) is true for all positive integers k. Principle of Mathematical Induction
  • 8. The assumption that P(k) is true is called the inductive hypothesis. When we use mathematical induction to prove a theorem, we first show that P(1) is true. Then we know that P(2) is true, because P(1) → P(2). Further, we know that P(3) is true, P(2) → P(3). Continuing along these lines, we see that P(n) is true for every positive integer n. Principle of Mathematical Induction
  • 9. 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we reach the next rung. Infinite Ladder
  • 10. 1. We can reach the first rung of the ladder. 2. If we can reach a particular rung of the ladder, then we reach the next rung. Basis Step: Verify that P(1) is true. Inductive Step: P(k) → P(k + 1) is true for all positive integers k Therefore, P(n) is true for all positive integers n, where P(n) is the statement that we can reach the nth rung of the ladder. Consequently, we can invoke mathematical induction to conclude that we can reach every rung. Infinite Ladder
  • 12. “Domino Effect” Step 1. The first domino falls Step 2. When any domino falls, the next domino falls
  • 13. “Domino Effect” Step 1. The first domino falls Step 2. When any domino falls, the next domino falls ….. all dominos will fall! This is how Mathematical Induction works.
  • 14. Example There are infinitely many stations on a train route. Suppose that the train stops at the first station and suppose that if the train stops at a station, then is stops at the next station. Show that the train stops at all stations.
  • 15. Solution Let P(n) be the statement that the train stops at station n. Basis Step: We are told that P(1) is true. Inductive Step: We are told that P(k) implies P(k+1) for each n ≥ 1. Therefore by the principle of the mathematical induction, P(n) is true for all positive integers n.
  • 16. Proving Summation Formulae Mathematical Induction is particularly well suited for proving that such formulae are valid. However, summation formulae can be proven in other ways. major disadvantage of using mathematical induction: 1. you cannot use it to derive this formula.
  • 17. Example Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2
  • 18. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Let P (n) be the proposition that the sum of the first n positive integers, 1 + 2 + ··· n = 𝑛 (𝑛+1) 2 , is 𝑛 (𝑛+1) 2 . We must do two things to prove that P (n) is true for n = 1, 2, 3,.... Namely, we must show that P(1) is true and that the conditional statement P(k)implies P(k+1) is true for k = 1, 2, 3,....
  • 19. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2
  • 20. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: We show that the conditional statement P(k) → P(k + 1) is true for all positive integers k.
  • 21. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: First, assume P(k) is true for all positive integers k.
  • 22. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: First, assume P(k) is true for all positive integers k. (Inductive Hypothesis)
  • 23. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Inductive Hypothesis: 1 + 2 + . . . + 𝑘 = 𝑘 (𝑘+1) 2
  • 24. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true.
  • 25. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true. 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘+1 [(𝑘+1)+1] 2 = 𝑘+1 (𝑘+2) 2
  • 26. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + . . . + 𝑘 = 𝑘 (𝑘+1) 2
  • 27. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) Left Side: P(𝑘): 1 + 2 + . . . + 𝑘 = 𝑘 (𝑘+1) 2
  • 28. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) Left Side: P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘 (𝑘+1) 2
  • 29. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) Right Side: P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘 (𝑘+1) 2 + (𝑘 + 1)
  • 30. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘 𝑘+1 + 2 𝑘+1 2
  • 31. Solution Show that if 𝑛 is a positive integer, then 1 + 2 + . . . + 𝑛 = 𝑛 (𝑛+1) 2 Basis Step: P(1) is true, because 1 = 1 (1+1) 2 Inductive Step: Add 𝑘 + 1 to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + . . . + 𝑘 + (𝑘 + 1) = 𝑘+1 𝑘+2 2
  • 32. Example Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛.
  • 33. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Let P(𝑛) be the proposition that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for the integer 𝑛.
  • 34. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1
  • 35. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Assume P(k) is true for an arbitrary nonnegative integer k.
  • 36. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Assume P(k) is true for an arbitrary nonnegative integer k. (Inductive Hypothesis)
  • 37. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Inductive Hypothesis: 1 + 2 + 22 + . . . + 2 𝑘 = 2 (𝑘 + 1) − 1
  • 38. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true.
  • 39. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Under this assumption, it must be shown that P(𝑘 + 1) is true. 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 1) + 1 −1 = 2 (𝑘 + 2) −1
  • 40. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 = 2 (𝑘 + 1) − 1
  • 41. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) Left Side: P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 1) − 1
  • 42. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) Right Side: P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 1) − 1 + 2 (𝑘 + 1)
  • 43. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 ⋅ 2 (𝑘 + 1) − 1
  • 44. Solution Use mathematical induction to show that 1 + 2 + 22 + . . . + 2 𝑛 = 2 (𝑛 + 1) − 1 for all nonnegative integers 𝑛. Basis Step: Let P(0) is true, because 20 = 1 = 2 1 − 1 Inductive Step: Add 2 (𝑘 + 1) to both sides of the equation in P(𝑘) P(𝑘): 1 + 2 + 22 + . . . + 2 𝑘 + 2 (𝑘 + 1) = 2 (𝑘 + 2) − 1
  • 45. 1. Prove that 12 + 32 + 52 + . . . + (2𝑛 + 1) 2 = ( 𝑛 + 1) (2 𝑛 + 1)(2 𝑛 + 3) 3 whenever 𝑛 is a nonnegative integer. 2. Prove that 3 + 3 ⋅ 5 + 3 ⋅ 52 + . . . + 3 ⋅ 5n = 3(5 𝑛 +1 − 1) 4 whenever 𝑛 is a nonnegative integer. Challenge (1/2 Sheet Paper)

Editor's Notes

  1. That is, we can show that P (n) is true for every positive integer n, where P (n) is the statement that we can reach the nth rung of the ladder.
  2. We assume that P(k)is true for an arbitrary positive integer k and show that under this assumption, P(k+1) must also be true.
  3. (The left-hand side of this equation is 1 because 1 is the sum of the first positive integer. The right-hand side is found by substituting 1 for n in n(n+1)/2.)
  4. This last equation shows that P(k + 1) is true under the assumption that P(k) is true. This completes the inductive step. We have completed the basis step and the inductive step, so by mathematical induction we know that P (n) is true for all positive integers n. That is, we have proven that 1 + 2 +···+n = n(n+1)/2 for all positive integers n.