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Using the principle of mathematical induction, prove that `(2^(3n)-1)` is divisible by `7` for all `n in Ndot`

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Let `P(n) : 2^(3n) - 1` is divisible by 7.
For `n = 1, 2^(3) - 1 = 7` , which is divisible by `7`.
Thus `P(1)` is true.
Let `P(n)` be true for some `n =k`.
Then `2^(3k) - 1 = 7m, m in N"………."(1)`
Now, `2^(3(k+1)) - 1 = 8 xx 2^(3k) - 1`
`= 8(7m+1)-1` [Using `(1)`]
`= 56m + 7`
`= 7(8m+1)`, which is divisible by 7.
Thus `P(k+1)` is true whenever `P(k)` is true.
So, by the principle of mathematical induction, `P(n)` is true for all natural numbers.
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