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Question
Prove by method of induction, for all n ∈ N:
(24n−1) is divisible by 15
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Solution
(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., 24k − 1 is a multiple of 15.
∴ 24k − 1 = 15a, where a ∈ N
∴ 24k = 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`
= 24k.24 − 1
= 16.(24k) − 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.
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