How to find upper and lower bound without using formula? I am studying discrete math for tomorrow's exam and got stuck in the below question. I tried to google it and couldn't find anything usefull.
Prove the following sum is theta(n^2) (we have to find O(n^2) and omega (n^2))
1) $P(n)= 1+2+3+4+\cdots +n$
2) $P(n) =n+(n+1)+(n+2)+\cdots +2n$
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.

For 1) if $n = 2k$, then $P(n) = P(2k) = 1+2+3+\cdots + 2k = (1+2k) +(2+ (2k-1)) + (3+(2k-2)) + \cdots + (k+(2k-(k-1))) = (2k+1) +(2k+1) + \cdots + (2k+1) = k(2k+1) = \dfrac{n}{2}\cdot (n+1) = \dfrac{n(n+1)}{2}$.
If $n = 2k+1$, then:
$1 + 2 + 3 +\cdots + (2k+1) = (1+2+3+\cdots + 2k) + (2k+1) = (1+2+\cdots + +(n-1)) + n = \dfrac{(n-1)n}{2} + n = \dfrac{n(n+1)}{2}$.
For 2) $n + (n+1) + (n+2) +\cdots + 2n = (n+n+n+\cdots + n) + (1+2+3+\cdots + n) = n\cdot n + \dfrac{n(n+1)}{2} = n^2 + \dfrac{n(n+1)}{2}$
