What is the total path length walked by a wood cutter to clear a square forest? I was watching this video on Youtube in which the youtuber calculates the total number of tiles walked by a villager in the game Age of Empires 2 to clear a 240 x 240 tile forest. The youtuber uses a numerical method to find the answer. 
However, I have another approach. In 1 dimension, the wood cutter walks 1 tile to the 1st tree, then walks back. Then 2 tiles to the second tree and walks back and lastly n tiles to the nth tree and walks n tiles back. This is the area of the triangle with $base = length / 2$ and $height = length / 2$. So, for $+x$ and $-x$ directions, the area should be $\frac{1}{2} \times length \times \frac{length}{2}$. Extending this intuition to 3 dimensions, we get the volume of a pyramid with $height = length/2$ and $base = length ^2 $ (its a square forest). 
$$\frac{1}{3} \times length ^ 2 \times \frac{length}{2} \times 2$$
When length is $240$, it gives $4,608,000$
 
Figure 1
The problem is that the answer calculated by the youtuber is $10,503,440$ which is off by a huge margin. Which one of us is correct?
 A: Where does he start?  It seems odd that he has to walk a full tile to the first tree.  It also seems odd that he can walk diagonally in a tile based game, so if he wants to go three tiles left and four down he just has to walk five instead of seven tiles.  
Based on the way you are counting the first row, the correct formula for clearing the first row is $\sum_{i=1}^{240}i=240 \cdot (240+1)$.  You said he goes out and back, which eliminates your factor $\frac 12$.  The sum of the numbers from $1$ to $n$ is $\frac 12n(n+1)$.  
If we imagine he starts in the tile left and next to the top left corner of the forest, which seems to match the first row, to clear the second row he walks $2\sqrt{1^2+1^2}$ to the first tree and back, $2\sqrt{2^2+1^2}$ for the second tree and back, and 2\sqrt{1^2+240^2} for the last tree and back.  The total distance is $2\sum_{i=1}^{240}\sqrt {1+i^2}$  
Adding over all the rows we have $$2\sum_{j=0}^{239}\sum_{i=1}^{240}\sqrt{i^2+j^2}\approx 21,156,300$$
according to Alpha.  
