11. Show that \( 9^{n+1}-8 n-9 \) is divisible by 64, whenever \( n \) is a positive integer.

To show that \( 9^{n+1} - 8n - 9 \) is divisible by 64 for any positive integer \( n \), we can use mathematical induction. ### Base Case: Let's start with the base case \( n = 1 \): \[ 9^{1+1} - 8 \cdot 1 - 9 = 9^2 - 8 - 9 = 81 - 8 - 9 = 64 \] Since \( 64 \) is divisible by \( 64 \), the base case holds. ### Inductive Step: Now, we assume that the statement is true for some positive integer \( k \), i.e., \[ 9^{k+1} - 8k - 9 \equiv 0 \mod 64 \] This means there exists an integer \( m \) such that: \[ 9^{k+1} - 8k - 9 = 64m \] We need to show that the statement holds for \( k + 1 \): \[ 9^{(k+1)+1} - 8(k+1) - 9 \] This simplifies to: \[ 9^{k+2} - 8(k+1) - 9 = 9^{k+2} - 8k - 8 - 9 = 9^{k+2} - 8k - 17 \] Now, we can express \( 9^{k+2} \) in terms of \( 9^{k+1} \): \[ 9^{k+2} = 9 \cdot 9^{k+1} \] Substituting this into our expression gives: \[ 9 \cdot 9^{k+1} - 8k - 17 \] Now, we can substitute \( 9^{k+1} \) from our inductive hypothesis: \[ 9 \cdot (8k + 64m + 9) - 8k - 17 \] Expanding this: \[ 9 \cdot 8k + 576m + 81 - 8k - 17 = 72k + 576m + 64 \] Now, we can factor out \( 64 \): \[ = 64 \cdot (9m + k + 1) \] Since \( 9m + k + 1 \) is an integer, we conclude that: \[ 9^{k+2} - 8(k+1) - 9 \equiv 0 \mod 64 \] ### Conclusion: By the principle of mathematical induction, we have shown that \( 9^{n+1} - 8n - 9 \) is divisible by \( 64 \) for all positive integers \( n \).