Unformatted text preview: 2 2 2 2 1 3" + "9 + 27 + . .. + 3k = 1 3k Â· With n = k + 1, the left side of the sum in question is 2 2 2 2 1 2 3" +"9 + 27 + . .. + 3k+l = (13k ) + 3k+ 1 3k+l 1 1 3k+ 1 = 1 3k+ 1 ' which is the desired formula. By the Principle of Mathematical Induction, the formula holds for all n ;::: 1. (g) [BB] When n = 1, the left side is 6  2 = 4 and the right side is 1(4), so the formula holds. Now assume that k ;::: 1 and that 4 + 10 + 16 + . .. + (6k 2) = k(3k + 1). We have 4 + 10 + 16 + . .. + [6(k + 1)  2] = 4 + 10 + 16 + . .. + (6k + 4) = (4 + 10 + 16 + . .. + (6k 2)) + (6k + 4) = k(3k + 1) + (6k + 4) using the induction hypothesis = 3k 2 + 7k + 4 = (k + 1)(3k + 4) which is the given formula with n = k + 1. By the Principle of Mathematical Induction, we conclude that 4 + 10 + 16 + . .. + (6n  2) = n(3n + 1) for all n ;::: 1....
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 Summer '10
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 Graph Theory, Mathematical Induction, Natural number, Mathematical logic, Mathematical proof, 1 3 k

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