Center of mass using vectors I encountered a physics problem in this book where I was given trees located at points $A,B,C,D,E$.
The problem tells me to move from $A$ to $B$, but only covering half the distance. From my current location, I would then move to $C$, but only covering one-third the distance. This pattern continues until, finally, I move to $E$, covering one-fifth the distance.
I was asked what would happen if the order of the trees was rearranged and the answer given was that I would get the same final point, which turned out to be the center of mass.
I'm failing to see how the procedure described in the problem is equivalent to averaging the $x$- and $y$-coordinates of the points.
I would really appreciate it if anyone could show how they are equivalent and why the order does not matter.
 A: Remember that the center of mass (assuming all trees have the same mass) is $\frac{1}{5}(A + B + C + D + E)$ (treating the locations as vectors).
After the first step, we end up with the point $\frac{1}{2}(A + B)$, found by taking their midpoint. Now from here, we travel $\frac{1}{3}$ of the distance to $C$. We take the weighted average to get the new position of
$$\frac{2}{3}\left(\frac{1}{2}(A + B)\right) + \frac{1}{3}C = \frac{1}{3}A +\frac{1}{3}B + \frac{1}{3}C$$
Similarly, after the next step, we end up at
$$\frac{3}{4}\left(\frac{1}{3}A +\frac{1}{3}B + \frac{1}{3}C\right) + \frac{1}{4}D = \frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4}D$$
Likewise, the final step brings us to
$$\frac{4}{5}\left(\frac{1}{4}A + \frac{1}{4}B + \frac{1}{4}C + \frac{1}{4}D\right) + \frac{1}{5}E = \frac{1}{5}A + \frac{1}{5}B + \frac{1}{5}C + \frac{1}{5}D + \frac{1}{5}E$$
Intuitively, each of the vectors adds less proportional weight each time, while those already used have their contributions slightly diminished each time. This has the effect of balancing out in the end to give the average.
