Prove the statement by the Principle of Mathematical Induction :4n �

1.	Prove the statement by the Principle of Mathematical Induction :4n – 1 is divisible by 3, for each natural number n.
Answer» According to the question, P(n) = 4ⁿ – 1 is divisible by 3. So, substituting different values for n, we get, P(0) = 4⁰ – 1 = 0 which is divisible by 3. P(1) = 4¹ – 1 = 3 which is divisible by 3. P(2) = 4² – 1 = 15 which is divisible by 3. P(3) = 4³ – 1 = 63 which is divisible by 3. Let P(k) = 4^k – 1 be divisible by 3, So, we get, ⇒ 4^k – 1 = 3x. Now, we also get that, ⇒ P(k+1) = 4^k+1 – 1 = 4(3x + 1) – 1 = 12x + 3 is divisible by 3. ⇒ P(k+1) is true when P(k) is true Therefore, by Mathematical Induction, P(n) = 4ⁿ – 1 is divisible by 3 is true for each natural number n.

Prove the statement by the Principle of Mathematical Induction :4n – 1 is divisible by 3, for each natural number n.

Answer»

According to the question,

P(n) = 4ⁿ – 1 is divisible by 3.

So, substituting different values for n, we get,

P(0) = 4⁰ – 1 = 0 which is divisible by 3.

P(1) = 4¹ – 1 = 3 which is divisible by 3.

P(2) = 4² – 1 = 15 which is divisible by 3.

P(3) = 4³ – 1 = 63 which is divisible by 3.

Let P(k) = 4^k – 1 be divisible by 3,

So, we get,

⇒ 4^k – 1 = 3x.

Now, we also get that,

⇒ P(k+1) = 4^k+1 – 1

= 4(3x + 1) – 1

= 12x + 3 is divisible by 3.

⇒ P(k+1) is true when P(k) is true

Therefore, by Mathematical Induction,

P(n) = 4ⁿ – 1 is divisible by 3 is true for each natural number n.