Answer:
[tex]f(n) = \frac35 5^n[/tex]
Step-by-step explanation:
Let's assume the solution is of the form [tex]f(n) = \alpha k^n[/tex], where [tex]\alpha[/tex] is to be determined on the starting condition, and [tex]k \in \mathbb{R} \ne 0[/tex]
[tex]f(n+1) -5f(n) = 0\\\alpha k^{n+1} -5 \alpha k^n=0\\\alpha k^n (k-5) = 0 \rightarrow k= 5[/tex]
at this point we can use our starting condition to determine the value of [tex]\alpha[/tex]:
[tex]f(1)=3 \rightarrow \alpha 5^1 =3 \rightarrow \alpha =\frac 35[/tex]
Our solution is, thus [tex]f(n) = \frac35 5^n[/tex]
