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

Using the principle ofmathematical induction, prove that `(2^(3n)-1)`i

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