Is my summation result correct? $\sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j) =\cdots= \frac{n(n-1)^2}{2}$

Question

Is this result correct?

$$\begin{align} \sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j) &= \sum_{i=0}^{n-1}\left( i^2 + \frac{i(i-1)}{2} \right)\\ &= \frac{3}{2}\sum_{i=0}^{n-1}i^2 - \frac{1}{2}\sum_{i=0}^{n-1}i \\ &= \frac{3}{2}\frac{n(n-1)(2n-1)}{2} + \frac{n(n-1)}{4} \\ &= \frac{n(n-1)^2}{2}. \end{align}$$

Answer 1

You are almost correct. It should be \begin{align} \sum_{i=0}^{n-1} \sum_{j=0}^{i-1} (i+j) &= \frac{3}{2}\sum_{i=0}^{n-1}i^2 - \frac{1}{2}\sum_{i=0}^{n-1}i \\ &= \frac{3}{2}\cdot \frac{n(n-1)(2n-1)}{\color{red}{6}} -\frac{1}{2}\cdot \frac{n(n-1)}{2}\\ &=\frac{n(n-1)((2n-1)-1)}{4}=\frac{n(n-1)^2}{2}. \end{align}