I understand the Big-O for this summation is $O(n^3)$. I tried to break it down algebraically and I seem to be getting $O(n^3):$
SUMMATION: \begin{align} \sum\limits_{i=1}^n i(n-i) &= \sum_{i=1}^n i \cdot \sum_{i=1}^n (n-i)\\ &= \sum_{i=1}^n i \cdot (\sum_{i=1}^n n- \sum_{i=1}^n i)\\ &= \frac{n(n+1)}2\cdot (n^2 \cdot n(n+1)/2) = O(n^4).\\ \end{align}
Might someone help point out my error here? Thanks!
Your first step is wrong: $\sum_{i=1}^ni(n-i)$ is not equal to $\left(\sum_{i=1}^ni\right)\left(\sum_{i=1}^n(n-i)\right)$. In fact
$$\begin{align*} \sum_{i=1}^ni(n-i)&=\sum_{i=1}^nin-\sum_{i=1}^ni^2\\ &=n\sum_{i=1}^ni-\frac{n(n+1)(2n+1)}6\\ &=\frac{n^2(n+1)}2-\frac{n(n+1)(2n+1)}6\\ &=\frac{3n^3+3n^2-(2n^3+3n^2+n)}6\\ &=\frac{n^3-n}6\\ &\in O(n^3)\;. \end{align*}$$