If f(n+1)+f(n-1)=2f(n) and f(0)=0 then f(n) is (here n in N, the set of natural numbers)

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If f(n+1)+f(n-1)=2f(n) and f(0)=0 then f(n) is (here n in N, the set of natural numbers)

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Valentina 8 months 2021-09-23T12:26:55+00:00 1 Answer 0 views 0

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    2021-09-23T12:28:26+00:00

    Answer:

    F(n) = 2 – 1/F(n-1) for all n greater than or equal to 2.

    So, F(2) = 2 – 1/F(1) = 2 – 1/3 = 5/3

    F(3) = 2 – 1/F(2) = 2 – 3/5 = 7/5

    F(4) = 2 – 1/F(3) = 2 – 5/7 = 9/7 and so on.

    Do you observe any pattern in the values? See if it matches with my observation:

    F(n) = (2n+1)/(2n-1), for all natural numbers n.

    (Plug in some values of n to verify!)

    or, F(n) = 1 + 2/(2n-1)

    Now,

    F(1) + F(2) + F(3) + • • • +F(1000)

    = 1000 + 2[ 1 + 1/3 + 1/5 + 1/7 + • • • + 1/1999]

    = 1000 – 2∑(1/2n-1), as n runs from 1 to 1000.

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