#

# Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality�) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.

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#

# Data Mining – Classification Prediction

There are two forms of data analysis that can be used for extracting models describing important classes or to predict future data trends. These two forms are as follows

Classification models predict categorical class labels; and prediction models predict continuous valued functions. For example, we can build a classification model to categorize bank loan applications as either safe or risky, or a prediction model to predict the expenditures in dollars of potential customers on computer equipment given their income and occupation.

## What is classification?

Following are the examples of cases where the data analysis task is Classification

A bank loan officer wants to analyze the data in order to know which customer (loan applicant) are risky or which are safe.

A marketing manager at a company needs to analyze a customer with a given profile, who will buy a new computer.

In both of the above examples, a model or classifier is constructed to predict the categorical labels. These labels are risky or safe for loan application data and yes or no for marketing data.

## What is prediction?

Following are the examples of cases where the data analysis task is Prediction

Suppose the marketing manager needs to predict how much a given customer will spend during a sale at his company. In this example we are bothered to predict a numeric value. Therefore the data analysis task is an example of numeric prediction. In this case, a model or a predictor will be constructed that predicts a continuous-valued-function or ordered value.

Note Regression analysis is a statistical methodology that is most often used for numeric prediction.

## How Does Classification Works?

With the help of the bank loan application that we have discussed above, let us understand the working of classification. The Data Classification process includes two steps

• Building the Classifier or Model
• Using Classifier for Classification

### Building the Classifier or Model

This step is the learning step or the learning phase.

In this step the classification algorithms build the classifier.

The classifier is built from the training set made up of database tuples and their associated class labels.

Each tuple that constitutes the training set is referred to as a category or class. These tuples can also be referred to as sample, object or data points.

### Using Classifier for Classification

In this step, the classifier is used for classification. Here the test data is used to estimate the accuracy of classification rules. The classification rules can be applied to the new data tuples if the accuracy is considered acceptable.

## Classification and Prediction Issues

The major issue is preparing the data for Classification and Prediction. Preparing the data involves the following activities

Data Cleaning Data cleaning involves removing the noise and treatment of missing values. The noise is removed by applying smoothing techniques and the problem of missing values is solved by replacing a missing value with most commonly occurring value for that attribute.

Relevance Analysis Database may also have the irrelevant attributes. Correlation analysis is used to know whether any two given attributes are related.

Data Transformation and reduction The data can be transformed by any of the following methods.

Normalization The data is transformed using normalization. Normalization involves scaling all values for given attribute in order to make them fall within a small specified range. Normalization is used when in the learning step, the neural networks or the methods involving measurements are used.

Generalization The data can also be transformed by generalizing it to the higher concept. For this purpose we can use the concept hierarchies.

Note Data can also be reduced by some other methods such as wavelet transformation, binning, histogram analysis, and clustering.

## Comparison of Classification and Prediction Methods

Here is the criteria for comparing the methods of Classification and Prediction

Accuracy Accuracy of classifier refers to the ability of classifier. It predict the class label correctly and the accuracy of the predictor refers to how well a given predictor can guess the value of predicted attribute for a new data.

Speed This refers to the computational cost in generating and using the classifier or predictor.

Robustness It refers to the ability of classifier or predictor to make correct predictions from given noisy data.

Scalability Scalability refers to the ability to construct the classifier or predictor efficiently; given large amount of data.

Interpretability It refers to what extent the classifier or predictor understands.

#

# Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality�) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.

#

# Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality�) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.

#

# Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality?) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.

#

# Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality?) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.

#

# Mathematical Induction – Problems With Solutions

The principle of mathematical induction is used to prove that a given proposition (formula, equality, inequality?) is true for all positive integer numbers greater than or equal to some integer N.

Let us denote the proposition in question by P (n), where n is a positive integer. The proof involves two steps:

Step 1: We first establish that the proposition P (n) is true for the lowest possible value of the positive integer n.

Step 2: We assume that P (k) is true and establish that P (k+1) is also true

Use mathematical induction to prove that

1 + 2 + 3 +. + n = n (n + 1) / 2

for all positive integers n.

Solution to Problem 1:

• Let the statement P (n) be

1 + 2 + 3 +. + n = n (n + 1) / 2

• STEP 1: We first show that p (1) is true.

Right Side = 1 (1 + 1) / 2 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 + 2 + 3 +. + k = k (k + 1) / 2

• and show that p (k + 1) is true by adding k + 1 to both sides of the above statement

1 + 2 + 3 +. + k + (k + 1) = k (k + 1) / 2 + (k + 1)

• The last statement may be written as

1 + 2 + 3 +. + k + (k + 1) = (k + 1)(k + 2) / 2

• Which is the statement p(k + 1).
• 1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 6

For all positive integers n.

Solution to Problem 2:

• Statement P (n) is defined by

1 2 + 2 2 + 3 2 +. + n 2 = n (n + 1) (2n + 1)/ 2

• STEP 1: We first show that p (1) is true.

Left Side = 1 2 = 1

Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1

• Both sides of the statement are equal hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 2 + 2 2 + 3 2 +. + k 2 = k (k + 1) (2k + 1)/ 6

• and show that p (k + 1) is true by adding (k + 1) 2 to both sides of the above statement

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = k (k + 1) (2k + 1)/ 6 + (k + 1) 2

• Set common denominator and factor k + 1 on the right side

= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6

• Expand k (2k + 1)+ 6 (k + 1)

= (k + 1) [ 2k 2 + 7k + 6 ] /6

• Now factor 2k 2 + 7k + 6.

= (k + 1) [ (k + 2) (2k + 3) ] /6

• We have started from the statement P(k) and have shown that

1 2 + 2 2 + 3 2 +. + k 2 + (k + 1) 2 = (k + 1) [ (k + 2) (2k + 3) ] /6

• Which is the statement P(k + 1).
• Use mathematical induction to prove that

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

for all positive integers n.

Solution to Problem 3:

• Statement P (n) is defined by

1 3 + 2 3 + 3 3 +. + n 3 = n 2 (n + 1) 2 / 4

• STEP 1: We first show that p (1) is true.

Left Side = 1 3 = 1

Right Side = 1 2 (1 + 1) 2 / 4 = 1

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

1 3 + 2 3 + 3 3 +. + k 3 = k 2 (k + 1) 2 / 4

• add (k + 1) 3 to both sides

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = k 2 (k + 1) 2 / 4 + (k + 1) 3

• factor (k + 1) 2 on the right side

= (k + 1) 2 [ k 2 / 4 + (k + 1) ]

• set to common denominator and group

= (k + 1) 2 [ k 2 + 4 k + 4 ] / 4

= (k + 1) 2 [ (k + 2) 2 ] / 4

• We have started from the statement P(k) and have shown that

1 3 + 2 3 + 3 3 +. + k 3 + (k + 1) 3 = (k + 1) 2 [ (k + 2) 2 ] / 4

• Which is the statement P(k + 1).
• Prove that for any positive integer number n. n 3 + 2 n is divisible by 3

Solution to Problem 4:

• Statement P (n) is defined by

n 3 + 2 n is divisible by 3

• STEP 1: We first show that p (1) is true. Let n = 1 and calculate n 3 + 2n

3 is divisible by 3

• hence p (1) is true.
• STEP 2: We now assume that p (k) is true

k 3 + 2 k is divisible by 3

is equivalent to

k 3 + 2 k = 3 M. where M is a positive integer.

• We now consider the algebraic expression (k + 1) 3 + 2 (k + 1); expand it and group like terms

(k + 1) 3 + 2 (k + 1) = k 3 + 3 k 2 + 5 k + 3

= [ k 3 + 2 k] + [3 k 2 + 3 k + 3]

= 3 M + 3 [ k 2 + k + 1 ] = 3 [ M + k 2 + k + 1 ]

• Hence (k + 1) 3 + 2 (k + 1) is also divisible by 3 and therefore statement P(k + 1) is true.
• Prove that 3 n n 2 for n = 1, n = 2 and use the mathematical induction to prove that 3 n n 2 for n a positive integer greater than 2.

Solution to Problem 5:

• Statement P (n) is defined by
• STEP 1: We first show that p (1) is true. Let n = 1 and calculate 3 1 and 1 2 and compare them
• 3 is greater than 1 and hence p (1) is true.
• Let us also show that P(2) is true.
• Hence P(2) is also true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by 3

3 * 3 k 3 * k 2

• The left side is equal to 3 k + 1. For k , 2, we can write

k 2 2 k and k 2 1

• We now combine the above inequalities by adding the left hand sides and the right hand sides of the two inequalities

2 k 2 2 k + 1

• We now add k 2 to both sides of the above inequality to obtain the inequality

3 k 2 k 2 + 2 k + 1

• Factor the right side we can write

3 * k 2 (k + 1) 2

• If 3 * 3 k 3 * k 2 and 3 * k 2 (k + 1) 2 then

3 * 3 k (k + 1) 2

• Rewrite the left side as 3 k + 1

3 k + 1 (k + 1) 2

• Which proves tha P(k + 1) is true
• Prove that n. 2 n for n a positive integer greater than or equal to 4. (Note: n! is n factorial and is given by 1 * 2 *. * (n-1)*n.)

Solution to Problem 6:

• Statement P (n) is defined by
• STEP 1: We first show that p (4) is true. Let n = 4 and calculate 4. and 2 n and compare them
• 24 is greater than 16 and hence p (4) is true.
• STEP 2: We now assume that p (k) is true
• Multiply both sides of the above inequality by k + 1

k! (k + 1) 2 k (k + 1)

• The left side is equal to (k + 1). For k , 4, we can write
• Multiply both sides of the above inequality by 2 k to obtain

2 k (k + 1) 2 * 2 k

• The above inequality may be written

2 k (k + 1) 2 k + 1

• We have proved that (k + 1)! 2 k (k + 1) and 2 k (k + 1) 2 k + 1 we can now write
• We have assumed that statement P(k) is true and proved that statment P(k+1) is also true.
• Use mathematical induction to prove De Moivre’s theorem

[ R (cos t + i sin t) ] n = R n (cos nt + i sin nt)

for n a positive integer.

Solution to Problem 7:

• STEP 1: For n = 1

[ R (cos t + i sin t) ] 1 = R 1 (cos 1*t + i sin 1*t)

• It can easily be seen that the two sides are equal.
• STEP 2: We now assume that the theorem is true for n = k, hence

[ R (cos t + i sin t) ] k = R k (cos kt + i sin kt)

• Multiply both sides of the above equation by R (cos t + i sin t)

[ R (cos t + i sin t) ] k R (cos t + i sin t) = R k (cos kt + i sin kt) R (cos t + i sin t)

• Rewrite the above as follows

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ (cos kt cos t – sin kt sin t) + i (sin kt cos t + cos kt sin t) ]

• Trigonometric identities can be used to write the trigonometric expressions (cos kt cos t – sin kt sin t) and (sin kt cos t + cos kt sin t) as follows

(cos kt cos t – sin kt sin t) = cos(kt + t) = cos(k + 1)t

(sin kt cos t + cos kt sin t) = sin(kt + t) = sin(k + 1)t

• Substitute the above into the last equation to obtain

[ R (cos t + i sin t) ] k + 1 = R k + 1 [ cos (k + 1)t + sin(k + 1)t ]

• It has been established that the theorem is true for n = 1 and that if it assumed true for n = k it is true for n = k + 1.