Chapter 4: Principle of Mathematical Induction

According to the question,

P(n) is 2n < (n + 2)!

So, substituting different values for n, we get,

P(0) ⇒ 0 < 2!

P(1) ⇒ 2 < 3!

P(2) ⇒ 4 < 4!

P(3) ⇒ 6 < 5!

Let P(k) = 2k < (k + 2)! is true;

Now, we get that,

⇒ P(k+1) = 2(k+1) ((k+1)+2))!

We know that,

But, we also know that,

= 2(k+1) × (k+3)(k+2)……………3×1 > 2(k+1)

Therefore, 2(k+1) < ((k+1) + 2)!

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 2n < (n + 2)! Is true for all natural number n.

According to the question,

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

√n < 1/√1 + 1/√2 + … 1/√n, for all natural numbers n ≥ 2

According to the question,

P(n) is 2 + 4 + 6 + …+ 2n = n² + n.

So, substituting different values for n, we get,

P(0) = 0 = 0² + 0 Which is true.

P(1) = 2 = 1² + 1 Which is true.

P(2) = 2 + 4 = 2² + 2 Which is true.

P(3) = 2 + 4 + 6 = 3² + 2 Which is true.

Let P(k) = 2 + 4 + 6 + …+ 2k = k² + k be true;

So, we get,

⇒ P(k+1) is 2 + 4 + 6 + …+ 2k + 2(k+1) = k² + k + 2k +2

= (k² + 2k +1) + (k+1)

= (k + 1)² + (k + 1)

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

2 + 4 + 6 + …+ 2n = n² + n is true for all natural numbers n.

According to the question,

P(n) is 1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1.

So, substituting different values for n, we get,

P(0) = 1 = 2⁰⁺¹ − 1 Which is true.

P(1) = 1 + 2 = 3 = 2¹⁺¹ − 1 Which is true.

P(2) = 1 + 2 + 2² = 7 = 2²⁺¹ − 1 Which is true.

P(3) = 1 + 2 + 2² + 2³ = 15 = 2³⁺¹ − 1 Which is true.

Let P(k) = 1 + 2 + 2² + … 2^k = 2^k+1 – 1 be true;

So, we get,

⇒ P(k+1) is 1 + 2 + 2² + … 2^k + 2^k+1 = 2^k+1 – 1 + 2^k+1

= 2×2^k+1 – 1

= 2^(k+1)+1 – 1

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1 is true for all natural numbers n.