(3n – 2n – 1) is divisible by 4 if and only if (3n – 2n – 1) is a multiple of 4.
Let P(n) ≡ (3n – 2n – 1) = 4m, where m ∈ N.
Step I: Put n = 1
∴ (3n – 2n – 1) = 3(1) – 2(1) – 1 = 0 = 4(0)
∴ (3n – 2n – 1) is a multiple of 4.
∴ P(n) is term for n = 1.
Step II:
Let us assume that P(n) is true for n = k.
∴ 3k – 2k – 1 = 4a, where a ∈ N
∴ 3k = 4a + 2k + 1 ….(i)
Step III:
We have to prove that P(n) is term for n = k + 1,
i.e., to prove that
3(k+1) – 2(k + 1) – 1 = 4b, where b ∈ N
P(k + 1) = 3k+1 – 2(k + 1) – 1
= 3k . 3 – 2k – 2 – 1
= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]
= 12a + 6k + 3 – 2k – 3
= 12a + 4k = 4(3a + k)
= 4b, where b = (3a + k) ∈ N
∴ P(n) is term for n = k + 1.
Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is term for all n ∈ N.
∴ 3n – 2n – 1 is divisible by 4, for all n ∈ N.