Sum of first n cubes - Mathematical Induction

In this video, I prove that the sum of the first n cubes is the square of the sum of the first n natural numbers

[richardleveson]

A concise presentation in clear unaffected English and a minimum of theatrics. Bravo!

[renesperb]

A very good example ! You do it very clearly in a nice handwriting.

[davidgagen]

If you simplify the RHS at the very start to [n(n+1)/2]^2 which of course becomes [k(k+1)/2]^2 then it becomes slightly easier to follow maybe since you can state at the WTS part that RHS needs to be [(k+1)(k+2)/2]^2 at that point. You know what ur chasing a bit sooner. Love these videos.

[cyruschang]

We need to show if 1^3 + 2^3 + 3^3 ... + k^3 = (1 + 2 + 3 ... + k)^2, then 1^3 + 2^3 + 3^3 ... + k^3 + (k + 1)^3 = (1 + 2 + 3 ... + k + (k + 1))^2
First we calculate the difference
(1 + 2 + 3 ... + k + (k + 1))^2 - (1 + 2 + 3 ... + k )^2 = 2(k + 1)(1 + 2 + 3 ... + k ) + (k+1)^2 = 2(k + 1)(k(1+ k)/2 ) + (k+1)^2 = k(1+ k)^2 + (k+1)^2 = (k+1)^2 (k + 1) = (k + 1)^3
Since (1 + 2 + 3 ... + k + (k + 1))^2 = (1 + 2 + 3 ... + k )^2 + (k + 1)^3
and
(1 + 2 + 3 ... + k )^2 = 1^3 + 2^3 + 3^3 ... + k^3
We showed
(1 + 2 + 3 ... + k + (k + 1))^2 = 1^3 + 2^3 + 3^3 ... + k^3 + (k + 1)^3

[RichardCorongiu]

You shoulda been a teacher...how relaxed and succinct...in other words.. a genius..nice work..
.

[Tairyokenois]

It's interesting. It works for other multiples as well (in the video's case, it's a series of multiples of 1) but we have to multiply by an additional value (in the video's case, it's x1, which is why it's not shown):
Here's a series of multiples of 7: 7^3 + 14^3 + 21^3 + 28^3 = (7+14+21+28)^2 x 7
Here's for 14: 14^3 + 28^3 + 42^3 + 56^3 + 70^3 + 84^3 = (14+28+42+56+70+84)^2 x 14
Here's for 166: 166^3 + 332^3 + 498^3 + 664^3 + 830^3 = (166+332+498+664 + 830)^2 x 166

In general it's: r^3 + (2r)^3 + (3r)^3 + (4r)^3 +...+ (nr)^3 = r(r + 2r + 3r + 4r +...+ nr)^2
where r is the common ratio and n is the number of terms. It only works if it follows specifically this general form, if it doesn't start at (r times 1)^3 then you'd have to subtract terms out from the answer.

[richardmullins]

looks very interesting!!!! congratultions on selecting this example !

[michaelbaum]

Very nice Induction Proof - 🙏 thanks a lot.

[downrightcyw]

Very nice mathematics teacher who will not blame or shout at students who got poor grade in Mathematics. Unlike our eastern strict Mathematics teachers.

[HeirTrap]

best handwriting ive seen in a while (30+years)!

[ralphw]

Thank you for this, I had seen the relationship and 
wondered how it was proved. Your presentations are superb.

[dronevluchten]

Wow. I have studied long ago math at the university of Utrecht (Netherlands) but this was unknown to me. 😊

[jorgepinonesjauch]

😮 se sabía que en un momento tenía que utilizar sumatorias conocidas, muy buena demostración!!

[niloneto]

Induction is easy. I wanna see someone proving identities like this and by deduction, as those expressions don't come from thin air.

[s.hariharan]

YOU BEATED MY MATHS TEACHER 😭, THANK YOU FOR GOOD PRESENTATION ❤❤...

[AZALI]

amazing video !!!
I'm very glad you covered this identity haha

coincidentally, I was just walking a tutee through this one a couple days ago !!

[Kid.Nimbus]

This is my first video I've seen of yours. Subscribed as soon as video was dond

[avalagum]

My idea before watching the video:
n = 1: correct
n = 2: correct
Suppose that 1^3 + ... + n^3 = (1 + ... + n)^2. We need to prove that 1^3 + ... + n^3 + (n + 1)^3 = (1 + ... + n + n+1)^2
(1 + ... + n + n+1)^2 = (1 + ... + n)^2 + 2(1 + ... + n)(n + 1) + (n + 1)^2
We need to prove that (n + 1)^3 = 2(1 + ... + n)(n + 1) + (n + 1)^2 which is easy to prove if we know 1 + ... + n = n(n + 1)/2

[tomdekler]

I doubt anyone at this level still needs to be explained why 1+2+3...+n = n(n+1)/2 but just in case.
First we write the series:
1+2+3...+n
Call this series p

Then beneath it we write the same series but backwards
n+n-1+n-2...1
Call this series q

Now we know both of these series have the same amount of terms, n of them.
So we can add these series term by term, and they will be equal to p+q.

Take the first two terms and add them, that's 1 + n
The second two terms added is 2 + n-1 = 1 + n
The third two terms added is 3 + n-2 = 1+n
And this pattern continues, all the way to the last terms that also add to 1+n

so you have n groups of terms that are all equal to (n+1), leading to a total addition of n(n+1)
You wrote the series twice, once forward and once backwards, and you added them.
Therefore, the sum of one of these series is n(n+1)/2

[qbfpojp]

U R a positive energy M. Teacher 🤩Merry Christmas