Answer: 199
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Step-by-step explanation:
In a prior answer, I mentioned how f(n) = f(n-1)+f(n-2) basically means "add the two terms prior to the nth term to get the nth term"
Let's use this idea to compute the third term:
Plug in n = 3
f(n) = f(n-1)+f(n-2)
f(3) = f(3-1)+f(3-2)
f(3) = f(2)+f(1)
f(3) = 1+2
f(3) = 3
Now plug in n = 4 to get the fourth term
f(n) = f(n-1)+f(n-2)
f(4) = f(4-1)+f(4-2)
f(4) = f(3)+f(2)
f(4) = 3+1
f(4) = 4
Repeat for n = 5. This process keeps going until we hit n = 12
f(n) = f(n-1)+f(n-2)
f(5) = f(5-1)+f(5-2)
f(5) = f(4)+f(3)
f(5) = 4+3
f(5) = 7
Plug in n = 6
f(n) = f(n-1)+f(n-2)
f(6) = f(6-1)+f(6-2)
f(6) = f(5)+f(4)
f(6) = 7+4
f(6) = 11
Plug in n = 7
f(n) = f(n-1)+f(n-2)
f(7) = f(7-1)+f(7-2)
f(7) = f(6)+f(5)
f(7) = 11+7
f(7) = 18
Plug in n = 8
f(n) = f(n-1)+f(n-2)
f(8) = f(8-1)+f(8-2)
f(8) = f(7)+f(6)
f(8) = 18+11
f(8) = 29
Plug in n = 9
f(n) = f(n-1)+f(n-2)
f(9) = f(9-1)+f(9-2)
f(9) = f(8)+f(7)
f(9) = 29+18
f(9) = 47
Plug in n = 10
f(n) = f(n-1)+f(n-2)
f(10) = f(10-1)+f(10-2)
f(10) = f(9)+f(8)
f(10) = 47+29
f(10) = 76
Plug in n = 11
f(n) = f(n-1)+f(n-2)
f(11) = f(11-1)+f(11-2)
f(11) = f(10)+f(9)
f(11) = 76+47
f(11) = 123
Plug in n = 12
f(n) = f(n-1)+f(n-2)
f(12) = f(12-1)+f(12-2)
f(12) = f(11)+f(10)
f(12) = 123+76
f(12) = 199
As you can see, it's a bit of work. Recursive sequences can be unwieldy.