5. A baker had 2 pans of corn bread. He served 1 1/4 pans. What fraction of a pan was left?
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Our function is 
.
We want to find f(5).
Let's substitute 5 for n.
f(5) = -2f(5-1) + 1
f(5) = -2f(4) + 1
We need to know f(4).
f(4) = -2f(4-1) + 1
f(4) = -2f(3) + 1
We need to know f(3).
f(3) = -2f(3-1) + 1
f(3) = -2f(2) + 1
We need to know f(2).
f(2) = -2f(2-1) + 1
f(2) = -2f(1) + 1
We know that f(1) = 3
f(2) = -2(3) + 1
f(2) = -6 + 1
f(2) = -5
Now we can use f(2) = -5 to find f(3)...
f(3) = -2(-5) + 1
f(3) = 10 + 1
f(3) = 11
Now we can use f(3) = 11 to find f(4)...
f(4) = -2(11) + 1
f(4) = -22 + 1
f(4) = -21
Now we can use f(4) = -21 t find f(5)!
f(5) = -2(-21) + 1
f(5) = 42 + 1
f(5) = 43
We want to find f(5).
Let's substitute 5 for n.
f(5) = -2f(5-1) + 1
f(5) = -2f(4) + 1
We need to know f(4).
f(4) = -2f(4-1) + 1
f(4) = -2f(3) + 1
We need to know f(3).
f(3) = -2f(3-1) + 1
f(3) = -2f(2) + 1
We need to know f(2).
f(2) = -2f(2-1) + 1
f(2) = -2f(1) + 1
We know that f(1) = 3
f(2) = -2(3) + 1
f(2) = -6 + 1
f(2) = -5
Now we can use f(2) = -5 to find f(3)...
f(3) = -2(-5) + 1
f(3) = 10 + 1
f(3) = 11
Now we can use f(3) = 11 to find f(4)...
f(4) = -2(11) + 1
f(4) = -22 + 1
f(4) = -21
Now we can use f(4) = -21 t find f(5)!
f(5) = -2(-21) + 1
f(5) = 42 + 1
f(5) = 43