Finding $f(x)$ where $x\in\mathbb{Z}$ such that $f(1)=2020$ and $\sum_{r=1}^{x}f(r)=x^2f(x)\forall x\gt 1$

Question

A function $f$ is defined for all positive integers and $f(1)=2020$ and $\sum_{r=1}^{x}f(r)=x^2f(x)\forall x\gt 1$. Find the value of $f(2019)$.

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My Attempt:

Replacing $x\mapsto x+1$. $f(x+1)=(x+1)^2f(x+1)-x^2f(x)$. This simplifies to $(x+2)f(x+1)=xf(x)$. This can be rewritten in the following form. $$\frac{f(x+1)}{f(x)}=\frac{x}{x+2}=1-\frac{2}{x+2}$$

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I am not sure how to solve this recurrence relation. Any hints are appreciated. Thanks

Answer 1

$\forall n\in\mathbb{N}^*,\frac{f(n+1)}{f(n)}=\frac{n}{n+2}$ thus $$ f(n)=f(1)\prod_{k=1}^{n-1}\frac{k}{k+2}=2020\prod_{k=1}^{n-1}\frac{k}{k+1}\prod_{k=1}^{n-1}\frac{k+1}{k+2}=2020\cdot\frac{1}{n}\cdot\frac{2}{n+1}=\frac{4040}{n(n+1)} $$

Answer 2

As you wrote, you have

$$\frac{x+2}{x}=\frac{f(x)}{f(x+1)}$$

or $$\frac{3}{1}=\frac{f(1)}{f(2)}$$ $$\frac{4}{2}=\frac{f(2)}{f(3)}$$ ...

$$\frac{2020}{2018}=\frac{f(2018)}{f(2019)}$$

now multiply and telescope.

Answer 3

As

$$ \sum_{r=1}^{x}f(r)=x^2f(x)\forall x\gt 1 $$

and

$$ \sum_{r=1}^{x+1}f(r)=(x+1)^2f(x+1)\forall x\gt 1 $$

subtracting the first from the second we have

$$ f(x+1) = (x+1)^2f(x+1)-x^2 f(x) $$

or

$$ f(x+1) = \frac{x}{x+2}f(x) $$

For $x\in\mathbb{Z}$ this Recurrence has the solution

$$ f(x) = \frac{C}{x(x+1)} $$