In my professors notes we have lemma 14 which states that $$\sum_{i=1}^{n} W(t_i)\sim\mathcal{N}\left(0,\sum_{i=1}^{n}(2n-(2i-1)t_i\right)$$ where $W(t)$ is the standard Brownian motion. The proof of this uses this identity without proving it $$\sum_{i,j}min(t_i,t_j) = \sum_{i=1}^{n}(2n-(2i-1))t_i$$
I want to try to prove this identity but I am confused on where to even start, any suggestions would be appreciated.
We assume that $t_1 < t_2 < \cdots < t_n$. Note that $\sum_{i=1}^n W(t_i)$ is normal with mean $0$ and variance \begin{align*} E\bigg(\Big(\sum_{i=1}^n W(t_i) \Big)^2 \bigg) &= \sum_{i, j=1}^n E\Big(W(t_i)W(t_j) \Big)\\ &= \sum_{i, j=1}^n \min(t_i, t_j)\\ &= \sum_{i=1}^n t_i + 2\sum_{i<j}\min(t_i, t_j)\\ &=\sum_{i=1}^n t_i + 2\sum_{i=1}^n (n-i)t_i\\ &= \sum_{i=1}^n\big(2n-2i +1 \big)t_i. \end{align*} That is, \begin{align*} \sum_{i=1}^n W(t_i) \sim \mathcal{N}\left(0, \ \sum_{i=1}^n\big(2n-2i +1 \big)t_i\right). \end{align*}