Question

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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

Answer:

$24(f(n))$

Step-by-step 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! :)