Prove the statement by the Principle of Mathematical Induction :For a

1.	Prove the statement by the Principle of Mathematical Induction :For any natural number n, 7n – 2n is divisible by 5.
Answer» According to the question, P(n) = 7ⁿ – 2ⁿ is divisible by 5. So, substituting different values for n, we get, P(0) = 7⁰ – 2⁰ = 0 Which is divisible by 5. P(1) = 7¹ – 2¹ = 5 Which is divisible by 5. P(2) = 7² – 2² = 45 Which is divisible by 5. P(3) = 7³ – 2³ = 335 Which is divisible by 5. Let P(k) = 7^k – 2^k be divisible by 5 So, we get, ⇒ 7^k – 2^k = 5x. Now, we also get that, ⇒ P(k+1)= 7^k+1 – 2^k+1 = (5 + 2)7^k – 2(2^k) = 5(7^k) + 2 (7^k – 2^k) = 5(7^k) + 2 (5x) Which is divisible by 5. ⇒ P(k+1) is true when P(k) is true. Therefore, by Mathematical Induction, P(n) = 7ⁿ – 2ⁿ is divisible by 5 is true for each natural number n.

Prove the statement by the Principle of Mathematical Induction :For any natural number n, 7n – 2n is divisible by 5.

Answer»

According to the question,

P(n) = 7ⁿ – 2ⁿ is divisible by 5.

So, substituting different values for n, we get,

P(0) = 7⁰ – 2⁰ = 0 Which is divisible by 5.

P(1) = 7¹ – 2¹ = 5 Which is divisible by 5.

P(2) = 7² – 2² = 45 Which is divisible by 5.

P(3) = 7³ – 2³ = 335 Which is divisible by 5.

Let P(k) = 7^k – 2^k be divisible by 5

So, we get,

⇒ 7^k – 2^k = 5x.

Now, we also get that,

⇒ P(k+1)= 7^k+1 – 2^k+1

= (5 + 2)7^k – 2(2^k)

= 5(7^k) + 2 (7^k – 2^k)

= 5(7^k) + 2 (5x) Which is divisible by 5.

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) = 7ⁿ – 2ⁿ is divisible by 5 is true for each natural number n.