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Answer :
We are given the recursive relation
[tex]$$
f(n+1) = \frac{1}{3} f(n)
$$[/tex]
and we know that
[tex]$$
f(3) = 9.
$$[/tex]
Since the relation shows that each term is one third of the previous term, we can work backwards to find the earlier terms.
1. Start with [tex]$f(3)$[/tex]:
The relation for [tex]$n=2$[/tex] is
[tex]$$
f(3) = \frac{1}{3} f(2).
$$[/tex]
To solve for [tex]$f(2)$[/tex], multiply both sides by 3:
[tex]$$
f(2) = 3 \cdot f(3) = 3 \cdot 9 = 27.
$$[/tex]
2. Next, find [tex]$f(1)$[/tex]:
The relation for [tex]$n=1$[/tex] is
[tex]$$
f(2) = \frac{1}{3} f(1).
$$[/tex]
Again, multiply both sides by 3 to isolate [tex]$f(1)$[/tex]:
[tex]$$
f(1) = 3 \cdot f(2) = 3 \cdot 27 = 81.
$$[/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 we know that
[tex]$$
f(3) = 9.
$$[/tex]
Since the relation shows that each term is one third of the previous term, we can work backwards to find the earlier terms.
1. Start with [tex]$f(3)$[/tex]:
The relation for [tex]$n=2$[/tex] is
[tex]$$
f(3) = \frac{1}{3} f(2).
$$[/tex]
To solve for [tex]$f(2)$[/tex], multiply both sides by 3:
[tex]$$
f(2) = 3 \cdot f(3) = 3 \cdot 9 = 27.
$$[/tex]
2. Next, find [tex]$f(1)$[/tex]:
The relation for [tex]$n=1$[/tex] is
[tex]$$
f(2) = \frac{1}{3} f(1).
$$[/tex]
Again, multiply both sides by 3 to isolate [tex]$f(1)$[/tex]:
[tex]$$
f(1) = 3 \cdot f(2) = 3 \cdot 27 = 81.
$$[/tex]
Thus, the value of [tex]$f(1)$[/tex] is
[tex]$$
\boxed{81}.
$$[/tex]
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