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Prove by the method of induction, for all n ∈ N.

3n – 2n – 1 is divisible by 4.

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(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.

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