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Question
Answer the following:
Prove by method of induction 152n–1 + 1 is divisible by 16, for all n ∈ N.
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Solution
152n–1 + 1 is divisible by 16, if and only if (152n–1 + 1) is a multiple of 16
Let P(n) ≡ 152n–1 + 1 = 16m, where m ∈ N.
Step I:
Put n = 1
∴ 152n–1 + 1 = `15^(2(1)–1)` + 1 = 15 + 1 = 16 = 16(1) which is a multiple of 16
∴ P(n) is true for n = 1
Step II:
Let us consider that P(n) is true for n = k
i.e., 152k–1 + 1 is a multiple of 16
∴ 152k–1 + 1 = 16a, where a ∈ N
∴ 152k–1 = 16a – 1
∴ `15^(2"k")/15` = 16a – 1
∴ 152k = 15.(16a – 1) ...(i)
Step III:
We have to prove that P(k + 1) is true
i.e., to prove that
`15^(2("k"+1)-1)` + 1 is a multiple of 16.
i.e., `15^(2("k"+1)-1)` + 1 = 16b, where b ∈ N
∴ `15^(2("k"+1)-1)` + 1
= 152k+1 + 1
= 152k .15 + 1
= 15(16a – 1) × 15 + 1 ...[From (i)]
= 225 × 16a – 225 + 1
= 225 × 16a – 224
= 16(225a – 14)
= 16 b, where b = (225a – 14) ∈ N.
∴ P(n) is true for n = k + 1
Step IV:
From all steps above by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 152n–1 + 1 is divisible by 16, for all n ∈ N.
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