Prove by induction that for all positive integers n , f(n) = 2^(3n+1) + 3*5^(2n+1) , is divisible by 17.

Prove the base caseFor n=0, f(0)= 2 + 15 = 17Therefore, when n=0, f(n) is divisible by 17, base case is true2. Assume true for any integerAssume for n=k, f(k) is divisible by 17f(k)= 2^3k+1 + 3(5^2k+1) ;3. Work out function for the next integerf(k+1) = 2^3k+4 + 3(5^2k+3) = 8(2^3k+1 + 3(5^2k+1)) + 2555^2k255 = 1517, therefore the second term is divisible by 172^3k+1 + 3(5^2k+1) = f(k), so if f(k) is divisible by 17, f(k+1) is divisible by 17.Since f(0) is divisible by 17, and if f(k) is divisible by 17, then f(k+1) is divisible by 17, then f(n) is divisible by 17 for all positive integers.

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