Find all sequences $f(0), f(1), f(2), . . .$ such that $∆f(x) = 2019f(x) + 2020$, where $x ∈ \mathbb N$

Question

I have this question which I'd appreciate some help on how to tackle it.

So far I've converted it into the recurrence relation $f(x+1) = 2020f(x) + 2020$

With the solution $f(x) = \alpha(2020)^x+\beta$, but I am unsure what the values of alpha and beta are.

Answer 1

Hint: You can use generating functions to solve given recurence. Look at this and this. In your case it will be $$ a_{n+1} = 2020 a_n + 2020 \\ ... \\ A(x) = 2020A(x) + \frac{2020}{1-x}$$ And that should be continued. At the finish think about that, how you can write your solution in terms of $f_0$.