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27 June, 10:17

Complete computefibonacci () to return fn, where f0 is 0, f1 is 1, f2 is 1, f3 is 2, f4 is 3, and continuing: fn is fn-1 + fn-2. hint: base cases are n = = 0 and n = = 1.

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  1. 27 June, 13:05
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    Fibonacci series is given by

    f (0) = 0, f (1) = 1, f (2) = 1, f (3) = 2, f (4) = 3 ...

    As you can see using the above result

    f (4) = f (3) + f (2),

    f (3) = f (2) + f (1),

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

    Proceeding using same pattern

    f (5) = f (4) + f (3) = 3+2=5

    f (6) = f (4) + f (5) = 3+5=8

    f (7) = f (6) + f (5) = 8+5=13

    f (8) = f (7) + f (6) = 13+8=21

    f (9) = f (8) + f (7) = 21+13=34

    f (10) = f (9) + f (8) = 34+21=55

    ...

    ...

    f (n-2) = f (n-3) + f (n-4)

    f (n-1) = f (n-2) + f (n-3)

    Similarly and finally we get using the same pattern we get, As given f (n) = f (n-1) + f (n-2).
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