Question

Prove by method of induction, for all n ∈ N:

(2⁴ⁿ−1) is divisible by 15

Answer 1

(2⁴ⁿ−1) is divisible by 15 if and only if (2⁴ⁿ−1) is a multiple of 15.

Let P(n) ≡ (2⁴ⁿ−1) = 15m, where m ∈ N

Step I:

Put n = 1

∴ 2⁴⁽¹⁾ − 1 = 16 − 1 = 15

∴ (2⁴ⁿ − 1) is a multiple of 15.

∴ P(1) is true

Step II:

Let us consider that P(n) is true for n = k

i.e., 2^4k − 1 is a multiple of 15.

∴ 2^4k − 1 = 15a, where a ∈ N

∴ 2^4k = 15a + 1    ...(i)

Step III:

We have to prove that P(k + 1) is true

i.e., to prove that `2^(4("k"+1)) − 1` is a multiple of 15.

i.e., `2^(4("k"+1)) − 1` = 15b, where b ∈ N

∴ P(k + 1) = `2^(4("k"+1)) − 1 = 2^(4"k"+4) − 1`

= 2^4k.2⁴ − 1

= 16.(2^4k) − 1

= 16(15a + 1) − 1   ...[From (i)]

= 240a + 16 − 1

= 240a + 15

= 15(16a + 1)

= 15b, where b = (16a + 1) ∈ N

∴ P(k + 1) is true

Step IV:

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

∴  (2⁴ⁿ − 1) is divisible by 15, for all n ∈ N.