How to find upper and lower bound without using formula I'm studying discrete math for tomorrow's exam and got stuck in the below question. I tried to google it and couldn't find anything useful.

Prove the following sum is $\Theta (n^2)$ (we have to find $O(n^2)$ and $\Omega (n^2)$)


$P(n)= 1+2+3+4\cdots + n$
$P(n) = n+(n+1)+(n+2)+\cdots +2n$


Note you cannot use any formula you have to do it my algebraic manipulation.
This might be simple but I am not getting any clue right now and I dont have solution of it.
Another approach:


$n^2/4=(n/2)^2= \underbrace{n/2 + n/2 +\cdots + n/2}_{n/2 \text{ times}} \le 1+2+\cdots + n \le \underbrace{n + n +\cdots + n}_{n \text{ times}} = n^2 $ 
$n^2= \underbrace{n + n +\cdots + n}_{n \text{ times}} \le n+(n+1)+\cdots + 2n \le \underbrace{2n + 2n +\cdots + 2n}_{n \text{ times}} = 2n^2 $