third term using the recursive rule f(1)=.2
for f(n)=2f(n-1)/5 + 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
To find the third term of the sequence defined by the given recursive rule, we can use the formula f(n) = 2f(n-1)/5 + 1.
Let's start by substituting n=1 into the formula to find the value of f(1):
f(1) = 2f(1-1)/5 + 1
f(1) = 2f(0)/5 + 1
However, we don't know the value of f(0) directly. To find it, we need to use the initial condition given in the question, f(1) = 0.2.
Substituting f(1) = 0.2, we have:
0.2 = 2f(0)/5 + 1
Now we can solve for f(0):
0.2 - 1 = 2f(0)/5
-0.8 = 2f(0)/5
To isolate f(0), we can multiply both sides of the equation by 5/2:
-0.8 * (5/2) = 2f(0)/5 * (5/2)
-4 = f(0)
Now that we have the value of f(0), we can go back to the original formula and substitute it in to find f(1):
f(1) = 2f(0)/5 + 1
f(1) = 2(-4)/5 + 1
f(1) = -8/5 + 1
f(1) = -8/5 + 5/5 = -3/5
We have found that f(1) = -3/5.
To find the third term, we can use the recursive rule again, substituting n=2:
f(2) = 2f(2-1)/5 + 1
f(2) = 2f(1)/5 + 1
f(2) = 2(-3/5)/5 + 1
f(2) = -6/25 + 1
f(2) = -6/25 + 25/25 = 19/25
Therefore, the third term of the sequence is 19/25.