It is a simple question not much backgrond information is required. However, I do need ton understand the proof of why.
Prove the finite affine space $(\mathbb{Z}/p)^3$, has only $p^2 +p +1$ lines passing through a point.
Also does this generalise for $(\mathbb{Z}/p)^n$ have a at most $p^{n-1}+p^{n-2}+\cdots+p+1$ lines passing through a point.
Look forward to your solutions.
Assuming you mean $\mathbb{Z}/p$.
Fix an element $x$. Then there are $p^3-1$ other points $y$ such that $xy$ is a line through $x$. But we have double-counted, because each line goes through other points too.
But of course, the points on each line are just $x$, $x+(x-y)=2x-y$, $3x-2y$, etc. until we get to $px-(p-1)y=y$ and we start to repeat. So each line contains at most $p-1$ distinct points (other than $x$). Are they all distinct? Well, if two of them are equal, then there must be $m,n \in \mathbb{Z}/p$ with $(m-n)(x-y) \equiv 0$, which can only happen if $m=n$ or $x=y$.
So there are $$\frac{p^3-1}{p-1}$$ distinct lines that pass through your given point. I'll leave it to you to see that that's equal to your formula, and how it generalises.