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3 August, 05:46

F (1) = 1 and f (n) = f (n-1) ^2+3

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

then find the value of

f (4)

+2
Answers (2)
  1. 3 August, 05:55
    0
    f (4) = 364

    Step-by-step explanation:

    f (1) = 1

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

    f (2) = f (1) ^2 + 3 = 1^2 + 3 = 4

    f (3) = f (2) ^2 + 3 = 4^2 + 3 = 16+3 = 19

    f (4) = f (3) ^2 + 3 = 19^2 + 3 = 361+3 = 364
  2. 3 August, 06:05
    0
    f (4) = 364

    Step-by-step explanation:

    f (n) = f (n-1) ² + 3

    Since f (1) = 1

    f (2) = [f (1) ]² + 3

    = 1² + 3

    = 4

    f (3) = [f (2) ]² + 3

    = 4² + 3

    = 19

    f (4) = [f (3) ]² + 3

    = 19² + 3

    = 364
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