Final answer:
To prove that 2n*3^(2n-1) is always divisible by 17, we can use proof by induction. In the base case, when n = 1, it leaves a remainder of 6. For the inductive step, assuming it holds for k, we need to prove it holds for k+1. By expressing it as 2k*3^(2k-1) + 2*3^(2k+1) and using the congruence of 9 ≡ -1 (mod 17), we can show that 2n*3^(2n-1) is divisible by 17 for all positive integers n.
Step-by-step explanation:
To prove that 2n*3^(2n-1) is always divisible by 17, we need to show that it leaves no remainder when divided by 17. We can use proof by induction to demonstrate this.
- Base case: When n = 1, 2n*3^(2n-1) = 2*3^1 = 6. 6 divided by 17 leaves a remainder of 6.
- Inductive step: Assume that for some positive integer k, 2k*3^(2k-1) is divisible by 17, i.e., it leaves no remainder.
- Show that it holds for k+1: We need to prove that 2(k+1)*3^(2(k+1)-1) is also divisible by 17. Let's express it as 2k*3^(2k-1) + 2*3^(2k+1). Since 2k*3^(2k-1) is divisible by 17 (by assumption), we just need to show that 2*3^(2k+1) is divisible by 17.
To prove that 2*3^(2k+1) is divisible by 17, we can write it as (2*9^k) * 3.
Since 9 is congruent to -1 modulo 17 (9 ≡ -1 (mod 17)), we can rewrite it as (-2*3^k) * 3 = -2*3^(k+1).
Therefore, we have shown that 2n*3^(2n-1) is divisible by 17 for all positive integers n, using proof by induction.