F(n) = f(n – 1) + f(n – 2); f(1) = 2; f(2) = 1.

find the 12th term.

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.

12r - 8 - 12

you have to combine like terms. below i have bolded the like terms:

12r - 8 - 12

together it combines to:

12r - 20

hope this !