Question

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

(2³ⁿ – 1) is divisible by 7

Answer 1

(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.