Consider the sets $A_1, A_2, \ldots, A_n$ where $|A_i|=k$, $n \geq k,$ and $|A_i \cap A_j| \leq 1$ for all $i,j$. What is the minimal numbers of elements in their union $|A_1 \cup A_2 \cup \ldots \cup A_n|?$
For $n=k$ I got that the union consist of $\frac{k(k+1)}{2}$ elements but what about the case $n>k?$
Also I have tried to use inclusion–exclusion formula to get any inequality but no success.
If $N$ is the number of elements in the union, then we have the lower bound $$ N(N-1) \ge n k(k-1). $$ To prove this, let $\deg x$ for an element $x \in A_1 \cup A_2 \cup \dots \cup A_n$ be the number of sets $A_i$ containing $x$. Since $$\sum_x \deg x = \sum_{i=1}^n |A_i| = nk,$$ the average value of $\deg x$ is $\frac{nk}{N}$, so there must be an element $x$ with $\deg x \ge \frac{nk}{N}$.
All the sets $A_i$ containing $x$ must contain $k-1$ other elements, with no repetition. So there are at least $1 + (k-1)\deg x$ elements in the union $A_1 \cup A_2 \cup \dots \cup A_n$. Using our inequality on $\deg x$, we get $$ 1 + (k-1) \frac{nk}{N} \le N \iff N(N-1) \ge nk(k-1). $$
This is going to be tight whenever $\deg x$ is constant (so that $\deg x = \frac{nk}{N}$ for all $x$) and the union of the sets containing $x$ is the same as the union of all the sets (in other words, for any element $y$ in the union, there is a set containing both $x$ and $y$).
In particular, if $k$ is a prime power, we can take a $d$-dimensional affine space over $\mathbb F_k$, which will have $N = k^d$ points and $n = \frac{k^d(k^d-1)}{k(k-1)}$ lines, giving us infinitely many values of $n$ at which the lower bound is tight.
(More generally, we can achieve the bound by an $(N,k,1)$-block design.)