I will solve a graph theory problem:
Deduce that $e(T_{n,k}) = \left \lfloor \left(\frac{k-1}{2k} \right) n^2\right \rfloor$ for all $k < 8$
I recommend that you read this post, first.
Graph Theory Problem Solving:
Since $k<8$, $-\frac{k}{8} > -1$. Therefore, $\frac{n^2(k-1)}{2k} – \frac{k}{8} >\frac{n^2(k-1)}{2k} -1$ and $ \left \lceil \frac{n^2(k-1)}{2k} – \frac{k}{8}\right \rceil \geq \left \lfloor \frac{n^2(k-1)}{2k} \right \rfloor$. Since $e(T_{n,k})$ is an integer, $ \left \lfloor \frac{n^2(k-1)}{2k} \right \rfloor \leq \left \lceil \frac{n^2(k-1)}{2k} – \frac{k}{8}\right \rceil \leq e(T_{n,k}) \leq \left \lfloor \frac{n^2(k-1)}{2k} \right \rfloor$. Therefore, $$e(T_{n,k}) = \left \lfloor \frac{n^2(k-1)}{2k} \right \rfloor$$.
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Source: Homework problem from Professor Choi‘s Graph Theory class, Gwangju Institute of Science and Technology, Fall Semester 2023