Asked by Anonymous
third term using the recursive rule f(1)=.2
for f(n)=2f(n-1)/5 + 1
for f(n)=2f(n-1)/5 + 1
Answers
Answered by
Reiny
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
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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