Answer :
We are given the recursive relation
[tex]$$
f(n+1)=\frac{1}{3} f(n)
$$[/tex]
and the value
[tex]$$
f(3)=9.
$$[/tex]
Since the recursion defines [tex]$f(n+1)$[/tex] in terms of [tex]$f(n)$[/tex], we can work backwards from [tex]$n = 3$[/tex] to find [tex]$f(1)$[/tex].
1. First, find [tex]$f(2)$[/tex]. The given formula can be rearranged in reverse. Since
[tex]$$
f(3)=\frac{1}{3} f(2),
$$[/tex]
multiplying both sides by [tex]$3$[/tex] gives
[tex]$$
f(2)=3 \times f(3)=3 \times 9=27.
$$[/tex]
2. Next, find [tex]$f(1)$[/tex] using the same reasoning. We have
[tex]$$
f(2)=\frac{1}{3} f(1),
$$[/tex]
and multiplying both sides by [tex]$3$[/tex] yields
[tex]$$
f(1)=3 \times f(2)=3 \times 27=81.
$$[/tex]
Thus, the value of [tex]$f(1)$[/tex] is [tex]$\boxed{81}$[/tex].
[tex]$$
f(n+1)=\frac{1}{3} f(n)
$$[/tex]
and the value
[tex]$$
f(3)=9.
$$[/tex]
Since the recursion defines [tex]$f(n+1)$[/tex] in terms of [tex]$f(n)$[/tex], we can work backwards from [tex]$n = 3$[/tex] to find [tex]$f(1)$[/tex].
1. First, find [tex]$f(2)$[/tex]. The given formula can be rearranged in reverse. Since
[tex]$$
f(3)=\frac{1}{3} f(2),
$$[/tex]
multiplying both sides by [tex]$3$[/tex] gives
[tex]$$
f(2)=3 \times f(3)=3 \times 9=27.
$$[/tex]
2. Next, find [tex]$f(1)$[/tex] using the same reasoning. We have
[tex]$$
f(2)=\frac{1}{3} f(1),
$$[/tex]
and multiplying both sides by [tex]$3$[/tex] yields
[tex]$$
f(1)=3 \times f(2)=3 \times 27=81.
$$[/tex]
Thus, the value of [tex]$f(1)$[/tex] is [tex]$\boxed{81}$[/tex].
