Stood out like a sore thumb to me... I mean... how do you end on such a simple mathematical error!? To be fair, the video wasn't riddled with mistakes (there were a couple), but... well, this kinda spoiled things.
Like he explained, the side length of the square constructed from all the unit cubes in the stack is the nth triangle number (1 + 2 + 3 + ... + n). As you can see, this is an arithmetic series from one to n. The sum is, of course, n(1 + n)/2 = (n^2 + n)/2. To get the total number of the unit cubes in the square (and in the whole stack), we need to square the nth triangle number. Therefore we get ((n^2 + n)/2)^2 = (n^4 + 2n^3 + n^2)/4. Edit: You don’t have to think of any arithmetic series at all. As he showed in the video, the nth triangle number can also be calculated using two converging triangles and dividing their combined area by two.
I wish you had done more than n=3 for the arbitrary case's visualization
The program is only slow because you print each result on the screen. Without that it takes no time.
6:13 it is 2n^3, not (2n)^3
Stood out like a sore thumb to me... I mean... how do you end on such a simple mathematical error!? To be fair, the video wasn't riddled with mistakes (there were a couple), but... well, this kinda spoiled things.
That was absolutely ridiculous. Thank you from Texas.
0:55 - but I'm feeling 22.
how did you derive that equation
Like he explained, the side length of the square constructed from all the unit cubes in the stack is the nth triangle number (1 + 2 + 3 + ... + n). As you can see, this is an arithmetic series from one to n. The sum is, of course, n(1 + n)/2 = (n^2 + n)/2. To get the total number of the unit cubes in the square (and in the whole stack), we need to square the nth triangle number. Therefore we get ((n^2 + n)/2)^2 = (n^4 + 2n^3 + n^2)/4.
Edit:
You don’t have to think of any arithmetic series at all. As he showed in the video, the nth triangle number can also be calculated using two converging triangles and dividing their combined area by two.
Doesn't this proof skip over the fact that there are infinitely many solutions where A is 0?
great!
The n must equal 1or2or3.