Selection Sort runtime in terms of Big O

I'm trying to understand why the sorting algorithm Selection Sort has asymptotic runtime in $O(n^2)$.

Looking at the math, the runtime is

$\qquad T(n) = (n-1) + (n-2) + \dots + 2 + 1$.

And this is stated to be equal to

$\qquad O(n^2)$.

However I just don't understand the intuition. I have tried several practical experiments for n=10 up to n=5000, and all point to that the time complexity of e.g. 5000 can never be greater T(5000) = 12.497.500 -- not T(5000) = 5000^2 = 25.000.000.

Now, I know that 5000 is not the same as infinity, but I just don't understand the intuition behind

$\qquad (n-1) + (n-2) + \dots + 2 + 1 = O(n^2)$.

Does someone have a great pedagogical explanation that my dim-witted mind can understand?

• 99% unrelated, 100% out of interest: why'd you use the word "pedagogical"? – enche Dec 3 '14 at 19:42
• I just wanted it explained it laymans terms that is understandable :) – Pætur Magnussen Dec 4 '14 at 23:32

Remember what $O(n^2)$ means: it means that there is a constant $c$ such that, for large enough $n$, the number of steps is at most $cn^2$. The constant doesn't have to be $1$; in this case, it looks like the constant is about $\tfrac12$.
Why is $(n-1) + (n-2) + \dots + 2+1$ equal to $O(n^2)$? Because you're adding up $n-1$ numbers and the average of those numbers is half-way between the first one and the last one, i.e., $\tfrac12(1 + (n-1))$. So the total is $$(n-1)\times \tfrac12(1 + (n-1)) = \tfrac12(n^2-n).$$Plug in $n=5000$ and you get... *drumroll*... $12497500$.
• @PæturMagnussen In this case, yes. But that's fine by the definition of $O(-)$. – David Richerby Dec 18 '13 at 18:44
• @Ben So substitute your favourite proof that $1 + \dots + k = \tfrac12k(k+1)$ and then substitute $k=n-1$. – David Richerby Dec 3 '14 at 0:17