Prove that "6^n + 9" is divisible by 5 for all natural numbers.
First assess that the initial case of where n = 1 is true. In this case, 6+9=15=53, so we can see that the case is true.We can then assume that 6k+9 is a multiple of 5, so we can let 6k+9 = 5A for some A in the natural numbers. We then consider the case of n = k+1, so consider 6k+1+96k+1+9 = 66k+9 = (6k+9) + (5*6k) = 5(A+6k) So it must be a multiple of 5The problem is shown true for the case of n = 1, and by assuming it is true for some k, it is shown to be true for the case n = k+1. So by the principle of mathematical induction it is true for all natural numbers n.
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