Using the principle to proof by mathematical induction we need to follow the techniques and steps exactly as shown.

We note that a prove by mathematical induction consists of three steps.

**• Step 1.** (Basis) Show that P(n₀) is true.

**• Step 2. ** (Inductive hypothesis). Write the inductive hypothesis: Let k be an integer such that k ≥ n₀ and P(k) be true.

**• Step 3. ** (Inductive step). Show that P(k + 1) is true.

Questions with solutions to Proof by Mathematical Induction

a + ar + ar

Let the given statement be P(n). Then,

P(n): a + ar + ar

When n = 1, LHS = a and RHS = {a(r

Therefore LHS = RHS.

Thus, P(1) is true.

Let P(k) be true. Then,

P(k): a + ar + ar

Now, (a + ar + ar

= a(r

Therefore,

P(k + 1): a + ar + ar

⇒ P(k + 1)is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

Proof by Mathematical Induction

(ab)

Let the given statement be P(n). Then,

P(n): (ab)

When = 1, LHS = (ab)

Therefore LHS = RHS.

Thus, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (ab)

Now, (ab)

= (a

= (a

= (a

Therefore P(k+1): (ab)

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

More examples to Proof by Mathematical Induction

Let the given statement be P(n). Then,

P(n): (x

When n = 1, the given statement becomes: (x

Therefore P(1) is true.

Let p(k) be true. Then,

P(k): x

Now, x

[on adding and subtracting x)

= x

⇒ P(k + 1): x

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

Hence, by the Principal of Mathematical Induction, P(n) is true for all n ∈ N.

Let P (n): (10

For n=1, the given expression becomes {10

So, the given statement is true for n = 1, i.e., P (1) is true.

Let P(k) be true. Then,

P(k): (10

⇒ (10

Now, {10

= 100 × {10

= (100 × 11m) - 99

= 11 × (100m - 9), which is divisible by 11

⇒ P (k + 1): {10

⇒ P (k + 1) is true, whenever P(k) is true.

Thus, P (1) is true and P(k + 1) is true , whenever P(k) is true.

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

Let P(n) : (7

For n = 1, the given expression becomes (7

So, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (7

⇒ (7

Now, {7

(on subtracting and adding 7 ∙ 3k)

= 7(7

= (7 × 4m) + 4 ∙ 3k

= 4(7m + 3

∴ P(k + 1): {7

⇒ P(k + 1) is true, whenever P(k) is true.

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

Solved examples to Proof by Mathematical Induction

(2 ∙ 7

Let P(n): (2 ∙ 7

For n = 1, the given expression becomes (2 ∙ 7

So, the given statement is true for n = 1, i.e., P(1) is true.

Let P(k) be true. Then,

P(k): (2 ∙ 7

⇒ (2 ∙ 7

Now, (2 ∙ 7

= (2 ∙ 7

= 7(2 ∙ 7

= (7 × 24m) - 6(5

= (24 × 7m) - 6 × 4p, where (5

[Since (5

= 24 × (7m - p)

= 24r, where r = (7m - p) ∈ N

⇒ P (k + 1): (2 ∙ 7

⇒ P(k + 1) is true, whenever P(k) is true.

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.

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

**●** **Mathematical Induction**

**Problems on Principle of Mathematical Induction**

**Proof by Mathematical Induction**

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