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Question
Prove by method of induction, for all n ∈ N:
3n − 2n − 1 is divisible by 4
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Solution
3n − 2n − 1 is divisible by 4 if and only if 3n − 2n − 1 is a multiple of 4.
Let P(n) ≡ 3n − 2n − 1 = 4m, where m ∈ N.
Step 1:
For n = 1, 3n − 2n − 1 = 3 − 2 − 1 = 0
which is divisible by 4.
∴ P(1) is true.
Step 2:
Let us assume that for some k ∈ N, P(k) is true,
i.e., 3k − 2k − 1 = 4a, where a ∈ N
∴ 3k = 4a + 2k + 1 ...(1)
Step 3:
To prove that P(k + 1) is true, i.e., to prove that
3k+1 − 2(k + 1) − 1 is a multiple of 4,
i.e., 3k+1 − 2(k + 1) − 1 = 4b, where b ∈ N
Now, 3k+1 − 2(k + 1) − 1 = 3k.3 − 2k − 2 − 1
= (4a + 2k + 1)3 − 2k − 3 ..[By (1)]
= 12a + 6k + 3 − 2k − 3
= 12a + 4k
= 4(3a + k)
= 4b, where b = (3a + k) ∈ 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., 3n − 2n − 1 is divisible by 4, for all n ∈ N.
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