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With induction we start with the base case n = 1. So setting n=1 we find that f(1) = 391 which is equal to 17x23. So indeed the base case holds.We assume that for positive integers k, f(k) is divisible by 17. We now seek to show that f(k+1) is also divisible by 17 and we can use the assumption that f(k) is. In this question it would seem smart to start with an expression of f(k+1).f(k+1) = 2^{3(k+1)+1} + 3(5^{2(k+1)+1})We will now try to manipulate this expression using index laws. We see that,f(k+1) = 2^{3k+4} + 3(5^{2k+3}) = 2^{3}2^{3k+1} + 3(5^{2}5^{2k+1}) = 8x2^{3k+1} + 3x25x5^{2k+1} = 8f(k) + 17x3(5^{2k+1})We have manipulated our expression to find that f(k+1) is divisible by 17 since we can assume f(k) is.So given any positive integer k we know that f(k+1) is divisible by 17. Since we showed in our base case that f(1) is divisible by 17 it follows that f(n) is divisible by 17 by induction for all positive integers n.

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