Question

Prove by 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 1:

For n = 1, 2³ⁿ − 1 = 2³ − 1 = 7, which is divisible by 7.

∴ P(1) is true.

Step 2:

Let us assume that for some k ∈ N, P(k) is true,

i.e., 2^3k − 1 = 7a, where a ∈ N

∴  2^3k = 7a + 1    ...(1)

Step 3:

To prove that P(k + 1) is true, i.e., to prove that

`2^(3("k"+1)) − 1` is a multiple of 7,

i.e., 2^3k+3 − 1 = 7b, where b ∈ N

Now, 2^3k+3 − 1 = 2^3k·2³ − 1

= (7a + 1)8 − 1    ...[By (1)]

= 56a + 7

= 7(8a + 1)

= 7b, where b = (8a + 1) ∈ N.

∴ P(k + 1) is true

Step 4:

From all the above steps and by the principle of mathematical induction, P(n) is true for all n ∈ N,

i.e., (2³ⁿ − 1)  is divisible by 7, for all n ∈ N.