Exercise 9 Page 447

We are asked to write the first $6$ terms of a sequence, given a recursive rule. To do so, we will use a table.

$n$	$f(n)=f(n-1)-f(n-2)$	$f(n-1)-f(n-2)$	$f(n)$
$0$	$f(0)=2$	$-$	$2$
$1$	$f(1)=4$	$-$	$4$
$2$	$f(2)=f(2-1)-f(2-2)$	$f(1)-f(0)=4-2$	$2$
$3$	$f(3)=f(3-1)-f(3-2)$	$f(2)-f(1)=2-4$	$\text{-}2$
$4$	$f(4)=f(4-1)-f(4-2)$	$f(3)-f(2)=\text{-}2-2$	$\text{-}4$
$5$	$f(5)=f(5-1)-f(5-2)$	$f(4)-f(3)=\text{-}4-(\text{-}2)$	$\text{-}2$

Therefore, the first $6$ terms of the sequence are $2,$ $4,$ $2,$ $\text{-}2,$ $\text{-}4$ and $\text{-}2.$