(3^{n} – 2n – 1) is divisible by 4 if and only if (3^{n} – 2n – 1) is a multiple of 4.

Let P(n) ≡ (3^{n} – 2n – 1) = 4m, where m ∈ N.

**Step I:** Put n = 1

∴ (3^{n} – 2n – 1) = 3^{(1)} – 2(1) – 1 = 0 = 4(0)

∴ (3^{n} – 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.

∴ 3^{k} – 2k – 1 = 4a, where a ∈ N

∴ 3^{k} = 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) = 3^{k+1} – 2(k + 1) – 1

= 3^{k} . 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.

∴ 3^{n} – 2n – 1 is divisible by 4, for all n ∈ N.