Question

Let the area bounded in the first quadrant by $\left[x\right]+\left[y\right]\le n,$ where $n\in R$ is denoted by $A\left(n\right)$. The value of $A\left(n\right)-A(n-1)-n$ is equal to
(Where [.] is greatest interger function)

Answer 1

When
$0\le y<1\Rightarrow \left[y\right]=0\Rightarrow \left[x\right]\le n\Rightarrow 0\le x<n+11\le y<2\Rightarrow \left[y\right]=1\Rightarrow \left[x\right]\le n-1\Rightarrow 0\le x<n$


Area is,
$1+2+3+\cdots +(n+1)\Rightarrow A\left(n\right)=\frac{(n+1)(n+2)}{2}\Rightarrow A(n-1)=\frac{\left(n\right)(n+1)}{2}\Rightarrow A\left(n\right)-A(n-1)=(n+1)\Rightarrow A\left(n\right)-A(n-1)-n=1$