A sequence is defined by the recursive function f(n + 1) = 1/3f(n). If f(3) = 9 , what is f(1)

Question

A sequence is defined by the recursive function f(n + 1) = 1/3f(n). If f(3) = 9 , what is f(1)

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Alice 2 weeks 2021-09-09T17:53:49+00:00 2 Answers 0

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    0
    2021-09-09T17:54:54+00:00

    Answer: D on edge 2020

    Step-by-step explanation:

    🙂

    0
    2021-09-09T17:55:20+00:00

    Answer:

    f(1) = 81

    Step-by-step explanation:

    given is f(n + 1) = 1/3f(n)

    f(3) = 9

    ayou need to find f(0) before you can answer what is f(1).

    Just start by applying for…

    n=0

    f(0+1) = 1/3 * f(0)

    f(1) = 1/3 * f(0)

    n=1

    f(1+1) = 1/3 * 1/3*f(0)

    f(2) = 1/9 * f(0)

    n=2

    f(2+1) = 1/3 * 1/9* f(0)

    f(3) = 1/27 * f(0)

    But it is given that f(3) = 9 so substitute this in the line above to calculate the value of f(0)….

    9 = 1/27 * f(0)

    Multiply left and right of the = sing by 27 gives

    f(0) = 27 * 9

    f(0) = 243

    We already know this next line because we started with n=0

    f(1) = 1/3 * f(0)

    Substitute f(0) = 243 in the line above to finally calculate the value of f(1).

    f(1) = 1/3 * 243

    f(1) = 81

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