Question

F(n+1)=5f(n), where f(1)=3 and n>1

Answer 1

`f(n) = \frac35 5^n`

Explanation:

Let's assume the solution is of the form
`f(n) = \alpha k^n`, where
`\alpha` is to be determined on the starting condition, and
`k \in \mathbb{R} \\e 0`

`f(n+1) -5f(n) = 0\\\alpha k^(n+1) -5 \alpha k^n=0\\\alpha k^n (k-5) = 0 \rightarrow k= 5`

at this point we can use our starting condition to determine the value of
`\alpha`:

`f(1)=3 \rightarrow \alpha 5^1 =3 \rightarrow \alpha =\frac 35`

Our solution is, thus
`f(n) = \frac35 5^n`