Unformatted text preview: 4k > 4Â· k4 by hypothesis. To prove the desired inequality 4 k + 1 > (k + 1)4 it is, therefore, sufficient to prove that 4Â· k4 ~ (k + 1)4. We know that (k + 1)4 = k4 + 4k 3 + 6k 2 + 4k + 1. Since 4 < k, 4k 3 < k4 and since 6 < k 2 , 6k 2 < k4. Also, 4k + 1 < 4k + k = 5k ::; kk < k4. Thus, (k + 1)4 < k4 + k4 + k4 + k4 = 4k4 as desired. 14. (a) [BB] Ifn = 0, (1 + v'2)O = 1 = 1 + 0v'2, so Xo = 1, Yo = O. If n = 1, (1 + v'2) I = 1 + v'2, so Xl = 1, YI = 1. If n = 2, (1 + v'2)2 = 1 + 2v'2 + 2 = 3 + 2v'2 so X2 = 3, Y2 = 2. If n = 5, (1 + v'2)5 = 1 + 5v'2 + 1O( v'2)2 + 10( v'2)3 + 5( v'2)4 + (v'2)5 1 + 5v'2 + 20 + 20v'2 + 20 + 4v'2 = 41 + 29v'2 and so X5 = 41, Y5 = 29. (b) We prove that, for n even, Xn 1 + 2(~) + 4(~) + 8(~) + . .. + 2n/2 Yn n + 2(;) + 4(~) + 8(~) + . .. + 2(n2)/2(n~l) while for n odd, Xn 1 + 2(~) + 4(~) + 8(~) + . .. + 2(nI)/2(n~l) Yn n + 2(;) + 4(~) + 8(~) + . .. + 2(nI)/2...
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 Summer '10
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 Graph Theory, Trigraph, xn yn, general term

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