good evening I want to show that: $\forall n \in \mathbb {Z}\quad :\quad n = E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right]$
$\begin{array}{l} m = E\left[ {\frac{{n - 1}}{2}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] \\ \forall k \in \mathbb {Z}:E\left[ {\frac{k}{2}} \right] + E\left[ {\frac{k}{2}} \right] = k \\ E\left[ {\frac{{n - 1}}{4}} \right] + E\left[ {\frac{{n - 1}}{4}} \right] \le E\left[ {\frac{{n - 1}}{2}} \right] \\ \Rightarrow 2E\left[ {\frac{{n - 1}}{4}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] \le m \\ \Rightarrow 4\left( {\frac{{n - 1}}{4}} \right) \le m \\ \Rightarrow n - 1 \le m.....\left( 1 \right) \\ m = E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 2}}{4} + \frac{1}{2}} \right] \\ E\left[ {\frac{{n + 4}}{4}} \right] = \left\{ \begin{array}{l} E\left[ {\frac{{n + 2}}{4}} \right] \\ E\left[ {\frac{{n + 2}}{4}} \right] + 1 \\ \end{array} \right. \\ m = E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + E\left[ {\frac{{n + 2}}{4}} \right] + E\left[ {\frac{{n + 4}}{4}} \right] \\ \Rightarrow m = \left\{ \begin{array}{l} E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + 2E\left[ {\frac{{n + 2}}{4}} \right] \\ E\left[ {\frac{{2\left( {n - 1} \right)}}{4}} \right] + 2E\left[ {\frac{{n + 2}}{4}} \right] + 1 \\ \end{array} \right. \\ \Rightarrow m < \left\{ \begin{array}{l} \frac{{2\left( {n - 1} \right)}}{4} + 2\frac{{n + 2}}{4} + 1 = n + 1 \\ \frac{{2\left( {n - 1} \right)}}{4} + 2\frac{{n + 2}}{4} + 2 = n + 2 \\ \end{array} \right. \\ \Rightarrow m < n + 1...\left( 2 \right) \\ \end{array}$
$ \begin{array}{l} E\left[ x \right] \\ x \le E\left[ x \right] < x + 1 \\ \end{array} $
disigne integral part How can we exclude the case $n-1$ thank you in advance