Final answer:
To prove that 7^2n - 1 is divisible by 48 for every natural number n, we can use induction. Similarly, to prove that 7^2n - 48n - 1 is divisible by 2304 for every natural number n, induction can be used as well.
Step-by-step explanation:
(a) To prove that 7^(2n) - 1 is divisible by 48 for every natural number n using induction, we start by checking the base case, which is n = 1. Plug in n = 1 to get 7^(2*1) - 1 = 48, which is divisible by 48. Next, assume that the statement is true for some arbitrary natural number k, which means 7^(2k) - 1 is divisible by 48. We need to show that it holds for k+1 as well. Consider 7^(2(k+1)) - 1 = 49 * (7^(2k) - 1) + 48. Since 7^(2k) - 1 is divisible by 48, and 49 is divisible by 48, we conclude that 7^(2(k+1)) - 1 is also divisible by 48. Therefore, the statement holds for all natural numbers n.
(b) To prove that 7^(2n) - 48n - 1 is divisible by 2304 for every natural number n using induction, we follow a similar process. Check the base case, which is n = 1. Plugging in n = 1, we get 7^2 - 48*1 - 1 = 48, which is divisible by 2304. Assume the statement is true for some natural number k, which means 7^(2k) - 48k - 1 is divisible by 2304. We need to prove it for k+1 as well. Consider 7^(2(k+1)) - 48(k+1) - 1 = 49 * (7^(2k) - 48k - 1) - 48. Since 7^(2k) - 48k - 1 is divisible by 2304 (by assumption), and 49 is divisible by 2304, we conclude that 7^(2(k+1)) - 48(k+1) - 1 is divisible by 2304. Hence, the statement holds for all natural numbers n.