F(n)=3^n find expressions for f(n+3) and f(n+1) hence find the value of k such that f(n+3)-f(n+1)= kf(n) where k ∈ N

Question

`f(n)=3^n`

find expressions for
`f(n+3)` and
`f(n+1)`
hence find the value of
`k` such that
`f(n+3)-f(n+1)= kf(n)` where k ∈ N

Answer 1

`24(f(n))`

Explanation:

`f(n+3) = 3^(n+3)\\f(n+1) = 3^(n+1)`

so

`f(n+3) - f(n+1) = 3^(n+3) - 3^(n+1)`

`= 27(3^n) - 3(3^n)`

`= 24(3^n)`

`= 24(f(n))`

hope this helped! :)