4 For every positive integer n, prove that 7"-3" is divisible by 4.

1.	4 For every positive integer n, prove that 7"-3" is divisible by 4.
Answer» Let P(n) : (7n– 3n) is divisible by 4. For n = 1, the given expression becomes (71- 31) = 4, which is divisible by 4. So, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, P(k): (7k- 3k) is divisible by 4. ⇒ (7k- 3k) = 4m for some natural number m. Now, {7(k + 1)- 3 (k + 1)} = 7(k + 1)– 7 ∙ 3k+ 7 ∙ 3k- 3(k + 1) (on subtracting and adding 7 ∙ 3k) = 7(7k- 3k) + 3k(7 - 3) = (7 × 4m) + 4 ∙ 3k = 4(7m + 3k), which is clearly divisible by 4. ∴ P(k + 1): {7(k + 1)- 3 (k + 1)} is divisible by 4. ⇒ 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.

Answer»

Let P(n) : (7n– 3n) is divisible by 4.

For n = 1, the given expression becomes (71- 31) = 4, which is divisible by 4.

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

Let P(k) be true. Then,

P(k): (7k- 3k) is divisible by 4.

⇒ (7k- 3k) = 4m for some natural number m.

Now, {7(k + 1)- 3 (k + 1)} = 7(k + 1)– 7 ∙ 3k+ 7 ∙ 3k- 3(k + 1) (on subtracting and adding 7 ∙ 3k)

= 7(7k- 3k) + 3k(7 - 3)

= (7 × 4m) + 4 ∙ 3k

= 4(7m + 3k), which is clearly divisible by 4.

∴ P(k + 1): {7(k + 1)- 3 (k + 1)} is divisible by 4.

⇒ 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.

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