Unformatted text preview: x + x 2 x 3 + ... ) i(1 + 3x + 5x 2 + 7x 3 + ... + (2n + l)xn + ... ) + ~(1 x + x 2 x 3 + ... ) Thus, an = i(2n + 1) + ~(_I)n. 13. (a) The characteristic polynomial of the Pell sequence is x 2 2x 1 whose roots are x = 1 ± .J2. Thus Pn = CI (1 + .J2)n + c2(1 .J2)n for constants CI and C2. Since Po = 1 and PI = 2, we have CI + C2 = 1 and CI (1 + .J2) + c2(1 .J2) = 2, so CI = 2+/'2 and C2 = 24 v'2. Thus Pn = (2+4v'2) (1 + .J2)n + (240) (1 .J2)n. (b) Note that 11 .J21 < ~, so 11 .J2l n < ~ for all n 2: 1. Also.J2 < 2 < 2 + .J2 implies o < 2 .J2 < 2, so 0 < 24 v'2 < ~ = ~. Thus the second term in the formula for Pn obtained in Exercise 13(a) is less than ~ for any n, that is, that IPn (2+4v'2) (1 + .J2)nl < ~ for any n. Since Pn is an integer, this part follows from the fact that there is precisely one integer within ~ of any given real number....
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 Summer '10
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 Graph Theory, Trigraph, pn, Xn

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