Summation of $O(n)$ from $1\le k\le n$
Suppose $T_n = O(n)$. That means that for some constant $C$, $T_n \leq Cn$. Therefore $$ \sum_{m=1}^n T_m \leq C \sum_{m=1}^n m = \frac{C}{2} m(m+1) = O(m^2). $$ If also $T_n = \Omega(n)$ then you can show in exactly the same way that the sum is $\Omega(n^2)$, so that no better bound is possible (for concreteness, you can take $T_n = n$).