A sequence is defined by the recursive function then f(1) is 81
Solution:
Given that,
A sequence is defined by the recursive function:
[tex]f(n+1) = \frac{1}{3} \times f(n)[/tex] ------ eqn 1
Given that,
[tex]f(3) = 9[/tex]
Substitute n = 2 in eqn 1
[tex]f(2 + 1) = \frac{1}{3} \times f(2)\\\\f(3) = \frac{1}{3} \times f(2)\\\\Substitute\ f(3) = 9\\\\9 = \frac{1}{3} \times f(2)\\\\f(2) = 27[/tex]
Substitute n = 1 in eqn 1
[tex]f(1 + 1) = \frac{1}{3} \times f(1)\\\\f(2) = \frac{1}{3} \times f(1)\\\\Substitute\ f(2) = 27\\\\27 = \frac{1}{3} \times f(1)\\\\f(1) = 27 \times 3\\\\f(1) = 81[/tex]
Thus f(1) is 81
