Unformatted text preview: 1 as required. By the Principle of Mathematical Induction, we conclude that the given assertion is true for all n 2: 1. (b) [BB] If n = 1,1 2 22 + 3 2 4 2 + . .. + (_I)n1 n2 = 12 = 1, by definition, while (_I)n1 n(n 2 +1) = (_1)111(1i 1 ) = 1, and so the result is true for n = 1. Now suppose that k 2: 1 and the result is true for n = k; that is, suppose that 12 _ 22 + 3 2 _ 42 + . .. + (_I)k1k2 = (ll1 k(k + 1). 2 We wish to prove that 12 22 + 3 2 4 2 + . .. + (_I)(k+1)1(k + 1)2 = (_1)(k+1)1 (k + 1)[(k + 1) + 1] 2 . Now 12 22 + 3 2 4 2 + . .. + (_I)(k+1)1(k + 1)2 = [12 _ 22 + 3 2 _ 4 2 + . .. + (_I)k1k2] + (_I)(k+1)1(k + 1)2 = (_I)k1 k(k + 1) + (I)k(k + 1)2 2 (using the induction hypothesis) = (_I)k1 k 2 + k + (_I)k1( 1)(k 2 + 2k + 1) 2 = (_I)k_1(k 2 ;k k 22kl) = (_I)k1 k 2 + k 2k2 4k  2 2 = (_I)k1 (k 2 2 3k  2) = (_I)k (k + 1)2(k + 2)...
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 Summer '10
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 Graph Theory, Mathematical Induction, Inductive Reasoning, Natural number, 1 K, Mathematical logic, Mathematical proof

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