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Unformatted text preview: 2 + . .. + (_1)n1 n 2 = L(1)k1 k 2 k=l n (c) 12 + 3 2 + 52 + . .. + (2n 1)2 = L(2k 1)2 k=l (d) 123 + 234 + 345 + . .. + n(n + l)(n + 2) n = L k(k + l)(k + 2) k=l 1 1 lIn 1 (e) +++ ... + = L . 12 23 34 n(n + 1) k=l k(k + 1) (Note that the last term on each of the left sides above provides a description of the general term required on the right. See the remarks following the expression (1).) 1 8. (a) [BB] For n = 1, L(i + 1)2i = 2(21) = 4, while n2 n + 1 = 1(22) = 4. Thus, the formula is i=l correct for n = 1. Now suppose that k 2': 1 and the formula is correct for n = k; thus, we suppose that k L(i + 1)2i = k2k+1. i=l We must prove that the formula is correct for n = k + 1; that is, we must prove that k+1 L(i + 1)2i = (k + 1)2k+2. i=l...
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 Summer '10
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 Graph Theory

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