Sum of cubes

In this video, I give an explicit formula for the sum of cubes, and I show in particular why it’s the square of the sum of integers. It is really clever and neat, enjoy!

[rnjsaudwo]

Oh my goodness... You solved a question that has puzzled me for long time ... I mean... ever since my high school days. Thank you very very much.!!!

[plaustrarius]

@0:43 So that means it's the square of the sum of n consecutive numbers!
Its also the triangular sum of odd numbers!


1 = 1^3
3 + 5 = 2^3
7+9+11= 3^3
etc.

So we can write it as an iterated sum!!

Edit: I now see BPRP has put up a video about iterated sum, it's like you guys plan this =D haha

[tur]

Wow this is a really clever way to get to that equality. I had no idea there could be such intuition behind this problem.

[TrimutiusToo]

You could have derived that area of the big square is actually (N*(N+1))^2 too... By just noticing that single outer side has N+1 squares of side length N
This would make proof even more awesome

[rome]

I got it as soon as you drew the squares. Thanks Dr Peyam.👍🏾👍🏾👍🏾

[AhmedHan]

Why is this channel not famous enough?

[JLConawayII]

This explanation is far better than mashing several Play-Doh cubes together to try to figure out what's going on.

[noniakishibusu]

4:10 "and 4 times N cubed... and gamecube"
A man with culture

[MrDannyDetail]

Wow, weirdly this video literally launched itself on my computer, I wasn't even using youtube at the time it suddenly started playing unbidden. But I love his enthusiasm for maths and will look out for more of his videos for sure.

[ed.puckett]

Thank you, this is so fascinating and beautiful!

I wanted to share my summary that helped me wrap my head around what your described:

4 1x1 squares arranged in a 2x2 square
8 2x2 squares surrounding the previous inner 2x2 square
12 3x3 squares surrounding the previous inner 6x6 square
16 4x4 squares surrounding the previous inner 12x12 square
20 5x5 squares surrounding the previous inner 20x20 square
...
4n nxn squares surrounding the previous inner 2Tnx2Tn square where Tn = n(n+1)/2 = n-th triangular number

Each successive layer consists of:
+ 4 nxn corners
+ n-1 nxn squares extending of each of the 4 sides
= (4 + 4*(n-1)) nxn squares = 4n nxn squares

Note: each side extension of each layer "fits" because it is subdividing (n-1)x(n-1) squares into n-1 pieces

[grgthbrg]

This is the most elegant proof of this identity I have seen yet!

[thomasborgsmidt]

Hmm... that brings back memories!
40 years ago I worked as a student in an exchange program in Germany and attended a party where I knew of Carl Friedrich Gauss that calculated the sum of the simple. The squares I also knew, but the cubes I did not. And then a passed a note that formulated the cube problem - with a questionmark. I'm not well trained in math.
It passed around till it got to a mathematician. Immediately he said: I don't know. But then:

1) Within a minute the boys were dispatched to get pen, paper and calculator. That taught me how organised Germans are - teamwork is their second nature.

2) All the girls threw their hands in the air and moaned - they KNEW that the boys would have no interest in them - before the problem was solved!

3) My contribution was that I remarked it was the simple squared.

4) But I never found the general formula.

[ulyssedurand]

Hello from France (thank you Dr Peyan, I love what you do, I am a highschool student, and when I can understand what you are doing I am very impressed)

[PackSciences]

When I learnt about this identity, I checked on the proofwiki "Sum of Sequence of Cubes
" and it turned out there were many different proofs for it.
You chose the visual demonstration but my favorite is the one by recursion.

[sea]

That proof was really cool! Added to my favourite videos list.

[bagusamartya]

Neat solving methods like this is one thing I love about math

[sergioh]

I always wondered the same thing thank you for clearing this up :)

[Fdan]

Great video. The algebra trick for the sum of all integers is so neat.

[MrCigarro]

Muchas gracias. En estadística cuando hacemos una transformación a rangos, esta suma se vuelve fundamental. Luego ser posible que mis alumnos puedan ver este vídeo es muy importante y motivante a la vez. Muchísimas gracias.

[AlwinMao]

0 dislikes 100% deserved so elegant so pleasurable