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Question
By the principle of mathematical induction, prove the following:
32n – 1 is divisible by 8, for all n ∈ N.
Solution
Let P(n) denote the statement 32n – 1 is divisible by 8 for all n ∈ N
Put n = 1
P(1) is the statement 32(1) – 1 = 32 – 1 = 9 – 1 = 8, which is divisible by 8
∴ P(1) is true.
Assume that P(k) is true for n = k.
i.e., 32k – 1 is divisible by 8 to be true.
Let 32k – 1 = 8m
To prove P(k + 1) is true.
i.e., to prove `3^(2(k+1)) - 1` is divisible by 8
Consider `3^(2(k+1)) - 1` = 32k+2 – 1
= 32k . 32 – 1
= 32k (9) – 1
= 32k (8 + 1) – 1
= 32k × 8 + 32k × 1 – 1
= 32k (8) + 32k – 1
= 32k (8) + 8m (∵ 32k – 1 = 8m)
= 8(32k + m), which is divisible by 8.
∴ P(k + 1) is true wherever P(k) is true.
∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.
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