Discrete Mathematics with Graph Theory (3rd Edition) 148

Discrete Mathematics with Graph Theory (3rd Edition) 148 -...

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Unformatted text preview: 146 Solutions to Exercises (b) -1,1,4,6,4, -4, -16, -24, -16, 16, .... The corresponding characteristic polynomial is x 2 - 2x + 2 whose roots are 1 i. By The- orem 5.3.1, the solution is an = cl(I + i)n + c2(1 - i)n. The initial conditions give ao = -1 = Cl + C2, al = 1 = cl(I + i) + c2(1- i), so Cl = -!(I + 2i), C2 = -!(I- 2i) and an = -!(I + 2i)(1 + i)n - !(I- 2i)(1- 2i)n. (c) -2,0,20, -40, -120,640, -80, -6240, 13280, 35840, .... The corresponding characteristic polynomial is x 2 + 2x + 10 whose roots are -1 3i. By Theorem 5.3.1, the solution is an = Cl (-1 + 3i)n + C2( -1 - 3i)n. The initial conditions give ao = -2 = Cl + C2, al = = Cl( -1 + 3i) + C2( -1 - 3i), so Cl = k( -3 + i), C2 = -k(3 + i) and an = k( -3 + i)( -1 + 3i)n - k(3 + i)( -1 - 3i)n. 21. [BB] In Problem 22, we showed that the nth term of the Fibonacci sequence is an-l = J-g () n - ....!...(l-y'S)n Thus la - ....!...(~)nl = 1....!...(l-y'S)nl Now 1....!...(l-y'S)nl < ....!... < 1 since y'S 2 . ,n-l y'S 2 y'S 2...
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