Consider the Random Walk $$X_t = X_{t-1}+\epsilon_t$$ with $\epsilon_t \sim N(\mu,\sigma^2)$ and $X_0=0$. We can write $$X_t=X_0+\sum_{n=1}^t\epsilon_t.$$ Using the equation above we have $$\mathbb{E}[X_t]=\mu t,$$ but I'm strugling to calculate the variance. I have \begin{equation} \begin{split} Var[X_t] &=\mathbb{E}[X_t^2]-\mathbb{E}[X_t]^2 \\ & = \mathbb{E} \left[\left(X_0+\sum_{n=1}^t\epsilon_t\right)^2\right]-(\mu t)^2 \\ & = \mathbb{E} [X_0^2]-2X_0\mathbb{E}\left[\sum_{n=1}^t\epsilon_t\right]+\mathbb{E}\left[\left(\sum_{n=1}^t\epsilon_t\right)^2\right] - (\mu t)^2 \end{split} \end{equation}
Well, now I think we can do $$\mathbb{E}\left[\left(\sum_{n=1}^t\epsilon_t\right)^2\right]=\sigma^2 t, \mathbb{E}[X_0^2]=0$$ and $$2X_0\mathbb{E}\left[\sum_{n=1}^t\epsilon_t\right]=2X_0\mu t=0$$ so we have $$Var(X_t) = \sigma^2 t - (\mu t)^2 .$$
But I'm not sure that this is right. Any help would be appreciated!
Assuming the $\epsilon_t$ are iid, you can simply use $$Var(X_t)=Var(\sum \epsilon_i)=\sum Var(\epsilon_i)$$