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20 April, 15:01

Find f (2), f (3), f (4), and f (5) if f is defined recur-sively by f (0) = f (1) = 1 and for n = 1, 2, ... a) f (n + 1) = f (n) - f (n - 1). b) f (n + 1) = f (n) f (n - 1). c) f (n + 1) = f (n) 2 + f (n - 1) 3. d) f (n + 1) = f (n) / f (n - 1). slader

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  1. 20 April, 16:30
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    (a) f (2) = 0, f (3) = - 1, f (4) = - 1, f (5) = 0

    (b) f (2) = 1, f (3) = 1, f (4) = 1, f (5) = 1

    (c) f (2) = 5, f (3) = 13, f (4) = 41, f (5) = 121

    (d) f (2) = 1, f (3) = 1, f (4) = 1, f (5) = 1

    Step-by-step explanation:

    (a)

    f (2) = f (1) - f (0) = 1 - 1 = 0

    f (3) = f (2) - f (1) = 0 - 1 = - 1

    f (4) = f (3) - f (2) = - 1 - 0 = - 1

    f (5) = f (4) - f (3) = - 1 - (-1) = 0

    (b)

    f (2) = f (1) * f (0) = 1*1 = 1

    f (3) = f (2) * f (1) = 1*1 = 1

    f (4) = f (3) * f (2) = 1*1 = 1

    f (5) = f (4) * f (3) = 1*1 = 1

    (c)

    f (2) = 2f (1) + 3f (0) = 2 + 3 = 5

    f (3) = 2f (2) + 3f (1) = 10 + 3 = 13

    f (4) = 2f (3) + 3f (2) = 26 + 15 = 41

    f (5) = 2f (4) + 3f (3) = 82 + 39 = 121

    (d)

    f (2) = f (1) / f (0) = 1/1 = 1

    f (3) = f (2) / f (1) = 1/1 = 1

    f (4) = f (3) / f (2) = 1/1 = 1

    f (5) = f (4) / f (3) = 1/1 = 1
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