Unformatted text preview: n ;::: 1. (d) When n = 1,1Â· 2Â·3+ 2Â·3Â·4+3Â·4Â·5+Â·Â·Â· +n(n+ l)(n+ 2) = 1Â· 2Â·3 = 6, by definition, while 1(1+1)(1:2)(1+3) = 1(2)~)(4) = 6, so the result holds. Now suppose that k ;::: 1 and the result is true for n = k; that is, suppose that 1.2.3+ 2.3.4 + ... + k(k + l)(k + 2) = k(k + l)(k: 2)(k + 3). We must prove that the result is true for n = k + 1, that is, that 1Â·2Â·3 + 2Â·3Â·4 + ... + (k + l)[(k + 1) + l][(k + 1) + 2J Now the left hand side of this equation is (k + l)[(k + 1) + l][(k + 1) + 2][(k + 1) + 3J 4 1Â·2Â·3+ 2 . 3 Â·4+ ... + (k + l)(k + 2)(k + 3) = [1.2Â·3 + 2Â·3Â·4 + ... + k(k + l)(k + 2) J + (k + l)(k + 2)(k + 3) = k(k + l)(k + 2)(k + 3) + (k + l)(k + 2)(k + 3) (using the induction hypothesis) 4 = (k + l)(k + 2)(k + 3) (~ + 1) = (k + l)(k + 2)1 k + 3)(k + 4) as required. By the Principle of Mathematical Induction, the given statement is true for all n ;::: 1....
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 Summer '10
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 Graph Theory, Mathematical Induction, Inductive Reasoning, Natural number, Mathematical logic, Mathematical proof

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