This is a question about the sum of an arithmetic sequence. Please forgive my lack of experience with LaTex.
To find the sum as asked, I wrote a general term for the arithmetic sequence and used the formula for the sum of the sequence, and the result was like this. $$\sum_{k=1}^n {ka_k} = \frac{n(n+1)}{2}\cdot\frac{(a_1+2a_n)}{3}$$ However, I think the fact that $\frac{a_1 + 2a_n}3$ is unique here. It's like an internal division formula. I don't think it's a coincidence, but I can't prove that it necessarily has to be like this. When $ka_k$ is listed from top to bottom, a triangle appears, and the coordinate of the triangle's center of gravity is $\frac{a_1 + 2a_n}3$, so I think it is related to this, but I could not develop my thoughts further. Could you please help?
The sum can be rewritten as
$$\begin{align} a_1+2a_2+3a_3+ \dots +na_n =\\ &+(a_1+a_2+\dots +a_n)\\ &+(a_2+a_3+\dots +a_n) \\ &+(a_3+a_4+\dots+a_n)\\ &+\dots \\ &+(a_{n-1}+a_n)\\ &+(a_n) \end{align}$$
These are $n$ sums of the last $k$ terms of the arithmetic sequence (where $k$ goes from $1$ to $n$).
Using the known formula for the sum of an arithmetic sequence, we get that our sum $S$ is
$$S= \sum _{k=1}^{n} \frac{n+1-k}{2}\cdot (a_{k}+a_n)=\frac{1}{2}\sum_{k=1}^{n}(n+1)a_n -ka_k +(n+1)a_k -ka_n$$
All of these terms are arithmetic sequences with respect to the index $k$, so we can simply sum them, except the term $ka_k$ which corresponds to our original sum $S$.
$$S=\frac{1}{2}(n(n+1)a_n-S+(n+1)\frac{n}{2}(a_1+a_n)-a_n\cdot \frac{n(n+1)}{2}) $$
Simplifying the equation for $S$, we get
$$3S=\frac{n(n+1)}{2}(2a_n+a_1+a_n-a_n)$$ $$S=\frac{n(n+1)}{2}\frac{2a_n+a_1}{3}$$
Which is what we were looking for.
As for why we are multiplying by the average of the corners of our "triangle", I could not think of a good explanation, since the center of mass relates to the average of the coordinates, not of the values themselves. However, There might be some sophisticated explanation relying on the constant difference between the terms.