Prove the statement the Principle of Mathematical Induction :For any

1.	Prove the statement the Principle of Mathematical Induction :For any natural number n, 7n – 2n is divisible by 5.
Answer» Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n. Now, P(1) = 7¹ - 2¹ = 5, which is divisible by 5. Hence, P(1) is true. Let us assume that, P(n) is true for some natural number n = k. .’. P(k) = 7^k -2^k is divisible by 5 or 7^k – 2^k = 5m, m ∈ N .......(i) Now, we have to prove that P(k + 1) is true P(k+ 1): 7^k+1 - 2^k+1 = 7^k.7 - 2^k.2 = (5 + 2)7^k - 2^k.2 = 5.7^k + 2.7^k -2^k.2 = 5.7^k + 2(7^k – 2^k) = 5 • 7^k + 2(5 m) (using (i)) = 5(7^k + 2m), which divisible by 5. Thus, P(k + 1) is true whenever P(k) is true. So, by the principle of mathematical induction P(n) is true for all natural numbers n.

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

Answer»

Let P(n): 7ⁿ – 2ⁿ is divisible by 5, for any natural number n.

Now, P(1) = 7¹ - 2¹ = 5, which is divisible by 5.

Hence, P(1) is true.

Let us assume that, P(n) is true for some natural number n = k.

.’. P(k) = 7^k -2^k is divisible by 5

or 7^k – 2^k = 5m, m ∈ N .......(i)

Now, we have to prove that P(k + 1) is true

P(k+ 1): 7^k+1 - 2^k+1

= 7^k.7 - 2^k.2

= (5 + 2)7^k - 2^k.2

= 5.7^k + 2.7^k -2^k.2

= 5.7^k + 2(7^k – 2^k)

= 5 • 7^k + 2(5 m) (using (i))

= 5(7^k + 2m), which divisible by 5.

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

So, by the principle of mathematical induction P(n) is true for all natural numbers n.