Let $t_i=\frac{T\cdot i}{n}$ for $T>0$, $i=1,...,n$ and let $(W_t)_{0\le t\le T}$ be a standard Brownian motion. Now I want to evaluate $$\text{var}\left(\sum_{i=1}^n W_{t_i}\right) = \mathbb E\left[\left(\sum_{i=1}^n W_{t_i}\right)^2\right]$$
I tried the following:
First of all, is this approach correct? But more importantly, is there a more intuitive/simpler way to tackle this problem?
(I apologise for having inserted a picture, but I didn't want to type all of this again.)
Yes, your approach is correct. Your calculation simplifies (a bit) if you use the symmetry, i.e. the fact that
$$\text{var} \left( \sum_{i=1}^n W_{t_i} \right) = \sum_{i=1}^n \text{var}(W_{t_i}) + 2 \sum_{i=1}^n \sum_{j=1}^{i-1} \text{cov}(W_{t_i},W_{t_j}).$$
An alternative approach is the following: The vector $(W_{t_1},\ldots,W_{t_n})$ is a Gaussian random variable with mean $0$ and covariance matrix $C=(t_j \wedge t_k)_{j,k=1}^n$. Since
$$\sum_{i=1}^n W_{t_i} = \underbrace{\begin{pmatrix} 1 & \ldots & 1 \end{pmatrix}}_{=:e^T} \cdot \begin{pmatrix} W_{t_1} \\ \vdots \\ W_{t_n} \end{pmatrix}$$
we know that $\sum_{i=1}^n W_{t_i}$ is Gaussian with mean $0$ and variance $e^T C e$. Consequently, we have to calculate $e^T C e$ and this leads to similar calculations.