(23n – 1) is divisible by 7 if and only if (23n – 1) is a multiple of 7.
Let P(n) ≡ (23n – 1) = 7m, where m ∈ N.
Step I:
Put n = 1
∴ 23n – 1 = 23(1) – 1 = 2 – 1 = 8 – 1 = 7
∴ (23n – 1) is a multiple of 7.
∴ P(n) is true for n = 1.
Step II: Let us assume that P(n) is true for n = k.
i.e., 23k – 1 is a multiple of 7.
∴ 23k – 1 = 7a, where a ∈ N
∴ 23k = 7a + 1 ……(i)
Step III: We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
23(k+1) – 1 = 7b, where b ∈ N.
∴ P(k + 1) = 23(k+1) – 1
= 23k+3 – 1
= 23k . (23) – 1
= (7a + 1)8 – 1 …..[From (i)]
= 56a + 8 – 1
= 56a + 7
= 7(8a + 1)
7b, where b = (8a + 1) ∈ N
∴ P(n) is true for n = k + 1.
Step IV: From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (24n – 1) is divisible by 7, for all n ∈ N.