For every positive integer n,P.T 7^n-3^n is divisible by 4
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Let P(n) : (7n – 3n) is divisible by 4.
For n = 1, the given expression becomes (7 1 - 3 1) = 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) + 3 k (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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For n = 1, the given expression becomes (7 1 - 3 1) = 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) + 3 k (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.
I hope it is helped you
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