Prove by the method of induction, for all n ∈ N.(23n – 1) is divis

1.	Prove by the method of induction, for all n ∈ N.(23n – 1) is divisible by 7
Answer» (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., 2^3k – 1 is a multiple of 7. ∴ 2^3k – 1 = 7a, where a ∈ N ∴ 2^3k = 7a + 1 ……(i) Step III: We have to prove that P(n) is true for n = k + 1, i.e., to prove that 2^3(k+1) – 1 = 7b, where b ∈ N. ∴ P(k + 1) = 2^3(k+1) – 1 = 2^3k+3 – 1 = 2^3k . (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.

Prove by the method of induction, for all n ∈ N.(23n – 1) is divisible by 7

Answer»

(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., 2^3k – 1 is a multiple of 7.

∴ 2^3k – 1 = 7a, where a ∈ N

∴ 2^3k = 7a + 1 ……(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

2^3(k+1) – 1 = 7b, where b ∈ N.

∴ P(k + 1) = 2^3(k+1) – 1

= 2^3k+3 – 1

= 2^3k . (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.