No idea how to solve SICP exercise 1.11

Question

Exercise 1.11:

A function f is defined by the rule that f(n) = n if n < 3 and f(n) = f(n - 1) + 2f(n - 2)

Answer 1

What did help me was running the process manually using a pencil and using hint author gave for the fibonacci example

a <- a + b
b <- a

Translating this to new problem is how you push state forward in the process

a <- a + (b * 2) + (c * 3)
b <- a
c <- b

So you need a function with an interface to accept 3 variables: a, b, c. And it needs to call itself using process above.

(define (f-iter a b c)
  (f-iter (+ a (* b 2) (* c 3)) a b))

If you run and print each variable for each iteration starting with (f-iter 1 0 0), you'll get something like this (it will run forever of course):

a   b   c
=========
1   0   0
1   1   0
3   1   1
8   3   1
17  8   3
42  17  8
100 42  17
235 100 42
...

Can you see the answer? You get it by summing columns b and c for each iteration. I must admit I found it by doing some trail and error. Only thing left is having a counter to know when to stop, here is the whole thing:

(define (f n)
  (f-iter 1 0 0 n))

(define (f-iter a b c count)
  (if (= count 0)
      (+ b c)
      (f-iter (+ a (* b 2) (* c 3)) a b (- count 1))))