Summation of asymptotic notation
You should be very careful when summing up a variable number of terms in asymptotic notation, as the result actually depends on the hidden constants.

Consider the following example: $f_i(n) = i\cdot n$ for all integers $i$ and $n$. Then, for any integer $i$, $f_i(n) \in O(n)$. 

If you are not careful, you could end up writing something like:
$$\sum_{i=1}^n f_i(n) = \sum_{i=1}^n O(n) = O(\sum_{i=1}^n n) = O(n^2).$$
And get $O(n^2)$. But this is totally wrong!

If you do the computation without the asymptotic notation, you get:
$$\sum_{i=1}^n f_i(n) = \sum_{i=1}^n i\cdot n = n\sum_{i=1}^n i = n\frac{n(n+1)}{2} = \frac{n^2(n+1)}{2}.$$
And this is not in $O(n^2)$, as it is in $\Theta(n^3)$.

Now some authors define $\sum_{i=1}^n O(n)$ as meaning that the hidden constants are the same for every $O$ and in this case you can sum things together. But this is not always the case so I recommend not using $O$ notation with variable-length sums (as well as when doing induction, where similar problems appear).