Use mathematical induction to prove that 7n+2 + 82n+1 is divisible by 57 for every nonnegative integer n.

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We need to prove that 7 to the power of n plus 2, plus 8 to the power of 2n plus 1, is divisible by 57 for every non-negative integer n. We'll use mathematical induction, which involves three main steps: establishing a base case, making an inductive hypothesis, and proving the inductive step. Let's start with the base case where n equals 0. We substitute n equals 0 into our expression: 7 to the power of 0 plus 2, plus 8 to the power of 2 times 0 plus 1. This gives us 7 squared plus 8 to the first power, which equals 49 plus 8, which equals 57. Since 57 is divisible by 57, our base case is proven. Now we state our inductive hypothesis. We assume that the statement is true for some non-negative integer k. This means we assume that 7 to the power of k plus 2, plus 8 to the power of 2k plus 1, is divisible by 57. In mathematical terms, this means the expression equals 57 times some integer m. For the inductive step, we need to prove the statement for n equals k plus 1. We start with 7 to the power of k plus 3, plus 8 to the power of 2k plus 3. We factor this as 7 times 7 to the power of k plus 2, plus 64 times 8 to the power of 2k plus 1. Using our inductive hypothesis, we substitute and simplify to get 57 times the quantity 7m plus 8 to the power of 2k plus 1, which is clearly divisible by 57. By the principle of mathematical induction, we have successfully proven our statement. We established the base case for n equals 0, made our inductive hypothesis, and proved the inductive step. Therefore, we can conclude that 7 to the power of n plus 2, plus 8 to the power of 2n plus 1, is divisible by 57 for every non-negative integer n. The proof is complete.