Show that :$$ \forall \in \mathbb{Z}, \left\lfloor \dfrac{n-1}{2}\right\rfloor +\left\lfloor \dfrac{n+2}{4}\right\rfloor + \left\lfloor\dfrac{n+4}{4} \right\rfloor =n$$
Solution provide by book:
In separate cases, as the remainder of the Euclidean division of $n$ by $4$ and present the results in a table
\begin{array} {|r|r|r|r|} \hline n & \left\lfloor \dfrac{n-1}{2}\right\rfloor & \left\lfloor \dfrac{n+2}{4}\right\rfloor & \left\lfloor\dfrac{n+4}{4} \right\rfloor & \sum \\ \hline 4k & 2k-1 & k & k+1 & 4k \\ \hline 4k+1 & 2k & k & k+1 & 4k+1\\ \hline 4k+2 & 2k & k+1 & k+1 & 4k+2 \\ \hline 4k+3 & 2k+1 & k+1 & k+1 & 4k+3 \\ \hline \end{array}
This establishes the desired result by examining all cases modulo $4$
- I didn't understand that solution which provided by Book could someone elaborate it Please ?
Reference
Because the least common multiple of the denominators is $4$, the effect of the floor functions will repeat with period $4$. You can convince yourself of this by trying $n$ from $0$ to $12$ or so. This means there are four cases depending on $n \bmod 4$ and they try all four cases. Each column in the middle of the table corresponds to one of the terms in the expression and they compute the value of that term,then add them up to get the last column. The fact that the first and last columns match establishes the identity. As and example, if $n=11,$ it is of the form $4k+3$ with $k=2$ Then $$\left\lfloor \dfrac{11-1}{2}\right\rfloor +\left\lfloor \dfrac{11+2}{4}\right\rfloor + \left\lfloor\dfrac{11+4}{4} \right\rfloor =5+3+3=11$$
Added per request: The point is that on adding one to nn exactly one of the floor functions will increase, maintaining the identity. Which one increases depends on $n \bmod 4$ Taking the LCM gets you through one cycle of the pattern.