algebra

third term using the recursive rule f(1)=.2

for f(n)=2f(n-1)/5 + 1

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  1. just keep working at it ...

    f(1) = .2
    f(2) = 2f(1)/5 + 1
    = 2(.2)/5 + 1
    = 1.08

    f(3) = 2f(2)/5 + 1
    = 2(1.08)/5 + 1
    = 1.432

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