Prove by the method of induction, for all n ∈ N.3n – 2n – 1 is d

1.	Prove by the method of induction, for all n ∈ N.3n – 2n – 1 is divisible by 4.
Answer» (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. ∴ 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. ∴ 3ⁿ – 2n – 1 is divisible by 4, for all n ∈ N.

Prove by the method of induction, for all n ∈ N.3n – 2n – 1 is divisible by 4.

Answer»

(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.

∴ 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.

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