For big-O complexity equations, how is something like $\frac{N(N-1)}{2}$ simplifies to just $O(N^2)$? I know the constants can be ignored because they are not significant, but when expanded,
For complexity equations, how is something like$$\frac{N(N-1)}{2} = \frac{N^2-N)}{2}$$
Only the $2$ is a constant, the other $N$ is not but why does it end up to only $O(N^2)$?

Question

For big-O complexity equations, how is something like $\frac{N(N-1)}{2}$ simplifies to just $O(N^2)$? I know the constants can be ignored because they are not significant, but when expanded, 
For complexity equations, how is something like$$\frac{N(N-1)}{2} = \frac{N^2-N)}{2}$$
Only the $2$ is a constant, the other $N$ is not but why does it end up to only $O(N^2)$?

anonymous · Answer

Look, we have :
$$\frac{ N^{2} - N }{ 2 } = \frac{ 1 }{ 2 } N^{2} - \frac{ 1 }{ 2 } N$$

So when we talk about the complexity, we usually ignore N because it's very very small than $$N^{2}$$
Now we have Complexity equal :
 $$\frac{ 1 }{ 2 } N^{2}$$

So with the O notation, the complexity will be : 
$$O(N^{2})$$

Good luck.