Principle Of Mathematical Induction

Let the given statement be P(n), as

Now for Proving a statement by mathematical Induction the steps involved are:
Step 1: Verify that P(1) is true.
Step 2: By taking P(k) as true prove that P(k + 1) is also true.

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

P(k + 1) = 1 + 3 + 3² .......+ 3^{[(k + 1) - 1]}

P(k + 1) = 1 + 3 + 3² ......+ 3^k

[From equation (1)]

Which is equal to the Right hand side for n = k + 1. We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1,

[As k + 2 = (k + 1) + 1]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

Steps for Proving a statement using Mathematical Induction are:
Step 1: Verify that P(1) is true.
Step 2: If P(k) is true then P(k + 1) is also true.

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right-hand side for n = k + 1. We proved that P(k + 1) is true.

Hence by the principle of mathematical induction, it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

Steps involved in Proving Mathematical Induction are:

Step 1: Verify that P(1) is true.

Step 2: If P(k) is true than P(k + 1) will also be true.

Step 1:

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right-hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by the principle of mathematical induction, it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1. We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1. We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1,

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Solving the left hand side with n = k + 1

[From equation (1)]

Which is equal to the Right hand side for n = k + 1.We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

…………..(1)

We shall prove that P(k + 1)is true,

Adding (k + 1) to both sides

We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

P(n):n(n + 1)(n + 5) is a multiple of 3.

For solving a statement P(n) by mathematical induction following steps are involved.

Step 1: Verify that P(1) is true.

Step 2: If P(k) is true then P(k + 1) is also true.

Step 1:

First, we check if it is true for n = 1,

P(1):1(2)(6) = 12 is a multiple of 3;

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

P(k): k(k + 1)(k + 5) = 3 m where m ∈ N ..........eq(1)

We shall prove that P(k + 1)is true,

P(k + 1): (k + 1)(k + 2)(k + 5 + 1)

⇒ (k + 1)(k + 2)(k + 5) + (k + 1)(k + 2)

⇒ k(k + 1)(k + 5) + (2)(k + 1)(k + 5) + (k + 1)(k + 2)

⇒ 3 m + (k + 1)[2 k + 10 + k + 2]

⇒ 3m + (k + 2)(3k + 12)

⇒ 3m + 3(k + 2)(k + 4)

⇒ 3[m + (k + 2)(k + 4)]

We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

is divisible by 11.

First, we check if it is true for n = 1,

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

P(k):10^{2k - 1} + 1 = 11m where m ∈ N

10^{2k - 1} = 11m - 1 ………….(1)

We shall prove that P(k + 1)is true,

P(k + 1):10^{2k + 1} + 1

⇒ 10^{2k - 1}.10² + 1

⇒ (11m - 1).100 + 1 From equation(1)

⇒ 1100m - 100 + 1

⇒ 1100m - 99

⇒ 11(100m - 9)

We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

P(n): x²ⁿ – y²ⁿ is divisible by (x + y).

First, we check if it is true for n = 1,

P(1): x² - y² = (x - y)(x + y);

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

P(k):x^2k - y^2k = m(x + y) where m ∈ N.

x^2k = y^2k + m(x + y) ………….(1)

We shall prove that P(k + 1) is true,

P(k + 1):x^{2k + 2} - y^{2k + 2}

⇒ x^2k.x² - y^{2k + 2}

⇒ [y^2k + m(x + y)]x² - y^{2k + 2} From equation(1)

⇒ m(x + y)x² + y^2k(x² - y²)

⇒ m(x + y)x² + y^2k(x - y)(x + y)

⇒ (x + y)[mx² + y^2k(x - y)]

We proved that P(k + 1) is true.

Hence by the principle of mathematical induction, it is true for all n ∈ N.

Let the given statement be P(n), as

P(n):3^{2n + 2} - 8n - 9 is divisible by 8.

First, we check if it is true for n = 1,

P(1):3⁴ - 8 - 9 = 81 - 17 = 64 = 8(8);

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

P(k):3^{2k + 2} - 8k - 9 = 8m where m ∈ N.

3^{2k + 2} = 8k + 9 + 8m ………….(1)

We shall prove that P(k + 1)is true,

P(k + 1):3^{2k + 4} - 8(k + 1) - 9

⇒ 3^{2k + 2}.3² - 8k - 8 - 9

⇒ (8k + 9 + 8m)9 - 8k - 17 From equation(1)

⇒ 64k + 72m + 81 - 17

⇒ 64k + 72m + 64

⇒ 8(8k + 9m + 8)

We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

P(n):41ⁿ - 14ⁿ is divisible by 27.

First, we check if it is true for n = 1,

P(1):41¹ - 14¹ = 27;

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

P(k):41^k - 14^k = 27m where m ∈ N.

41^k = 14^k + 27m ………….(1)

We shall prove that P(k + 1)is true,

P(k + 1):41^{k + 1} - 14^{k + 1}

⇒ 41^k.41 - 14^{k + 1}

⇒ (14^k + 27m)41 - 14^{k + 1} From equation(1)

⇒ 27.41m + 14^k(41 - 14)

⇒ 27.41m + 14^k.27

⇒ 27(41m + 14^k)

We proved that P(k + 1) is true.

Hence by principle of mathematical induction it is true for all n ∈ N.

Let the given statement be P(n), as

P(n):(2n + 7) < (n + 3)²

Step 1:

First, we check if it is true for n = 1,

P(1): (2 + 7) < (4)²;

∴ It is true for n = 1.

Now we assume that it is true for some positive integer k, such that

P(k):(2k + 7) < (k + 3)² .......(1)

We shall prove that P(k + 1)is true,

⇒ (2k + 7) + 2 < (k + 3)² + 2

⇒ 2(k + 1) + 7 < k² + 6k + 11

⇒ 2(k + 1) + 7 < k² + 6k + 11 + (2k + 5)

⇒ 2(k + 1) + 7 < k² + 8k + 16

⇒ 2(k + 1) + 7 < (k + 4)²

We proved that P(k + 1) is true.

Hence by the principle of mathematical induction, it is true for all n ∈ N.