Justification for expression for Convex Hull

I am new to convex hulls and I have encountered the following statement:

Given a set of n points in a vector space $$\{x_1 ... x_n \}$$, every point $$x_0$$ in the convex hull they form is obtained from the expression:

$$x_0 = \sum_{i=1}^n \alpha_i x_i$$ where $$\alpha_i\geq0$$ and $$\sum_{i=1}^n\alpha_i=1$$, with different coefficients $$\alpha_i$$ giving different points, always within the convex hull.

I have seen in this other answer a very clear example in $$\mathbb R^2$$ with only three points, for which the convex hull is just the convex combination of those three points, and corresponds to the points within the triangle formed by them. In $$\mathbb R^2$$, but for $$n$$ points, the convex hull is just the set of all points inside the polygon formed by the "outer" points of the set - enclosing the rest of "inner" points. (Formally, the smallest convex set which contains all the points in our set of $$n$$ points.)

If we added another point inside the triangle, the convex hull would remain the same; however, based on the expression I provide, that new point should also be included in the weighed sum (the sum goes from $$0$$ to $$n$$), in spite of lying inside such polygon and therefore not altering the convex hull. Why would such inner points be included in the sums? Are perhaps their coefficients zero (i.e. $$\alpha_i=0$$ precisely for those "inner" points)?

Additionally, if possible, I would like to know how the expression I give for the most general case is obtained, or at least how I could intuitively think about it to make it a bit more transparent.

When you include points in the interior of the convex hull in the defining set then the representation will not be unique. Different sets of coefficients can represent the same point. In fact an interior point in the defining set can be represented with its coefficient $$1$$ and the others $$0$$, and also represented as a convex combination of the vertices of the convex hull.