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

[@kangsungho1752]

Beauty of Mathmatical Induction

[@richardleveson6467]

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.

[@davidgagen9856]

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.

[@stephenlesliebrown5959]

Another superb presentation 😊

[@JohnDoe-jj6yd]

Nice & elegant! Congrats, my friend👍

[@michaelbaum6796]

Very nice Induction Proof - 🙏 thanks a lot.

[@HeirTrap]

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

[@jorgepinonesjauch8023]

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

[@Kid.Nimbus]

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

[@user-kj7hr3qw9w]

Amazing Prof 🎉

[@cyruschang1904]

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

[@mohamedlekbir6086]

Bravo professeur. J'aime bien.

[@ralphw622]

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. 😊

[@user-kw5qv6zl5e]

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

[@wouterzoons1843]

That was beautiful!

[@user-qb8fp8oj1p]

U R a positive energy M. Teacher 🤩Merry Christmas

[@ignantxxxninja]

Nice proof! Also nice handwriting

[@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.

[@gilbertoamigo7205]

Fantastic! Tanks, teacher.

[@thopita]

Beautiful problem ❤

[@s.hariharan6958]

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

[@iwallcool3377]

very clear. Bravo!

[@romaobraz4295]

wow that was beautiful

[@raminrasouli191]

Thank you very much. You are great.

[@tuanmanhtoan]

Very nice

[@vaibhavsrivastva1253]

Impressive!

[@niloneto1608]

Induction is easy. I wanna see someone proving identities like this and 1²+2²+3²+...+n²=n(n+1)(2n+1)/6 by deduction, as those expressions don't come from thin air.

[@achomik1999]

They do: Let f(x)=e^x + e^2x + e^3x +...+ e^nx. Geometric sequence, so f(x)=e^x*(e^nx - 1)/(e^x - 1). f(x)=e^nx - 1 + (e^nx - 1)/(e^x - 1) What interests us is the limit of k-th derivative of f(x) as x->0. Examples: Sum of n 1's (1=1⁰=2⁰=n⁰; k=0) is lim as x->0 of e^x*(e^nx - 1)/(e^x - 1)=lim as x->0 of n*e^nx/e^x=n*e^(0*n)/e^0=n*1/1=n (way to define natural numbers?) It requires more calculations for bigger k.

[@achomik1999]

@@samueldeandrade8535 Yes. k-th derivative of f(x)=e^x + e^(2x) +...+e^(nx) is e^x + (2^k)e^(2x) + (3^k)e^(3x) +...+ (n^k)e^(nx); plug in x=0 to get 1+2^k+3^k+...+n^k. f(x) can be written as (e^x)(1-e^(nx))/(1-e^x), here you cannot plug in x=0 but we can look for the limit as x->0 using d'Hospital's rule.

[@achomik1999]

@@samueldeandrade8535 Yeah, quite cool. If one integrated the right side of the equation, they would get stuff like 1+1/2+1/3+...+1/n or even the Riemann Zeta function in a limit.

[@shadrana1]

If you use Difference Theory these formulae are easy to prove.Note my post on this current proof. After you do the difference spadework, You arrive at a=1/3,b=1/2,c= 1/6 and d=0. S2(n)=1^2+2^2+3^2+..........+n^2= n^3/3+n^2/2+n/6+0*d =(2n^3+3n^2+n)/6 =n(2n^2+3n+1)/6 =n(n+1)(n+2)/6 which is the formula we need QED. for the S4(n) you will need to solve a 5X5 simultaneous group. The hard bit is building up and solving the equations without making mistakes.

[@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.

[@TakeAbackPak]

Wonderful!

[@AZALI00013]

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 !!

[@user-gz7tx8ur7r]

Wonderful

[@Calcprof]

My favorite surprising result easily proved by induction.

[@khl0513]

You inspire me, great!

[@PrimeNewtons]

Wow, thank you!

[@belhajabdellah2047]

Tanks. Very good !

[@benmooiman1174]

Mooi bewijs door twee keer toe te passen: 1^2 +2^2 + … n^2 = n * (n +1) / 2. Fraai!! 👍

[@thuongcap2]

Chứng minh bằng phương pháp quy nạp. Hay!

[@X-Joker7]

Love from India 🇮🇳❤

[@vitotozzi1972]

Awesome!!!!

[@tarciso21claudia28]

Maravilhoso !!!

[@baselinesweb]

Wpi;d it be incorrect to treat both sums as integrals and then equate? The you get n^4/4=(n^2/2)^2. This seems to prove it. What am I missing?

[@tomiokashw]

Perfect!

[@tobyfitzpatrick3914]

Pure Poetry..!

[@abdoulayesow6627]

That's clean as a proof.

[@AH-jt6wc]

can you please explain at which moment you verify the assomption that the proposition is true for n=k ? I mean everything is based on that. What if it is wrong for a "random" k ?

[@PrimeNewtons]

You don't. Because it was true for the first few tries, you assume it is true for n=k. If your assumption is false, mathematical induction would fail.

[@douwevandermaden2736]

i love it

[@Grassmpl]

You can derive such a formula using telescoping sums (i+1)^4-i^4.

[@joseantoniodominguezllover7774]

Excelente

[@avalagum7957]

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

[@jasdlf]

Always note that “it is true for all positive real numbers”!!

[@gaiatetuya92]

嬉しそうな良い顔してるねえ。

[@michaelhargus4316]

Interesting

[@binyaminaharoni1743]

Awesome lesson. The guy is also very cute. 😅

[@PrimeNewtons]

Thanks

[@Perception-2310]

thank you air

[@tim_cleezy]

where did the cube go

[@oneli8492]

掌握技巧是狭隘的，掌握方法是有限的，掌握原理才是终极道理。

[@NickEdgington]

nice

[@shadrana1]

You can use difference theory to prove this. (1) S1(n)= 1^1+2^1+3^1+..........+n^1= n(n+1)/2 (n is a member of natural numbers) (2) S2(n)= 1^2+2^2+3^2+..........+n^2=n(2n+1)(2n+2)/6 (3) S3(n)= 1^3+2^3+3^3+..........+n^3= ((n)(n+1)/2))^2 = (S1(n))^2 (4) S4(n)= 1^4+2^4+3^4+..........+n^4= n(n+1)(2n+1)(3n^2+3n-1)/30. Let S3(n)= 1^3+2^3+3^3+............+n^3= (n(n+1)/2)^2= (1+2+3+.......+n)^2 S3(6)=1^3+2^3+3^3+4^3+5^3+6^3 1+8+27+64+125+216= (n(n+1)/2)^2=441.........................(1) S3(1)=1^3=1 S3(2)=1^3+2^3=9 S3(3)=36,S3(4)=100,S3(5)=225,S3(6)=441 1st difference= 8,27,64,125,216. 2nd difference=19,37,61,91. 3rd difference=18,24,30. 4th difference=6,6 In S3(n), n>6 the 4th difference will always be 6 no matter the magnitude of n. Therefore the difference table will always work for n. This suggests S3(n) is a fourth order polynomial, S3(n)=an^4+bn^3+cn^2+dn+e...............................................(2) S3(1)=1=a+b+c+d+e..............................................................(3) S3(2)=9=16a+8b+4C+2d+e...................................................(4) S3(3)=36=81a+27b+9c+3d+e...............................................(5) S3(4)=100=256a+64b+16c+4d+e........................................(6) S3(5)=225=625a+125b+25c+5d+e......................................(7) S3(6)=441=1296a+216b+36c+6d+e....................................(8) Sweep from (3)-(8), 8 =15a+7b+3c+d...................................................................(9) 27=65a+19b+5c+d.................................................................(10) 64=175a+37b+7c+d...............................................................(11) 125=369a+61b+9c+d.............................................................(12) 216=671a+91b+11c+d..........................................................(13) Sweep from (9)-(13), 19=50a+12b+2c....................................................................(14) 37=110a+18b+2c..................................................................(15) 61=194a+24b+2c..................................................................(16) 91=302a+30b+2c..................................................................(17) Sweep from (14)-(17), 18=60a+6b............................................................................(18) 24=84a+6b............................................................................(19) 30=108a+6b..........................................................................(20) Sweep away the bs, 6=24a, >>>>>>>>>>>>>a=1/4,b=1/2,c=1/4,d=0,e=0 Going back to (2), S3(n)=n^4/4+n^3/2+n^2/4=(n^4+2n^3+n^2)/4=(n^2(n+1)^2)/2^2=((n(n+1))/2)^2 =(1^3+2^3+3^3+.................+n^3)=(1+2+3.............+n)^2 case proved. S3(6)=1^3+2^3+3^3+4^3+5^3+6^3 = 441 = (1+2+3+4+5+6)^2)= (21)^2 =441 in the S3(6) particular case. Thanks for the brilliant example of mathematical induction. Well done young man.

[@tomdekler9280]

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

[@Kid.Nimbus]

This shit fires me up i love math

[@darcash1738]

Why does this happen?

[@user-ig7nj1xb5r]

3: 50 the assume is sure? by what ?

[@roger7341]

It is common knowledge that 1+2+...+n=n(n+1)/2 and 1^3+3^3+...+n^3=n^2(n+1)^2/4. Thus, [n(n+1)/2]^2=n^2(n+1)^2/4, and equality is verified.

[@skwbusaidi]

Fir more calrification , when we get tge base case n=1 is true And assume that for n=k>=1 is true and found that it is true k+1. We done and some might say why. I tell them that if we assume that it is true for k and find out that it is true for k+1 Then we go back to base case k=1 If it is true for n=1 , it is true fir n=2" And if true for n=2, it true for n=3 And we go on

[@chengkaigoh5101]

Can we prove why induction works as a proof?

[@keescanalfp5143]

no, we can't . we just have to trust the so called 'farmer's logica'. the matter is : first: you show that the thesis is true for let's say n = 1 , or like this video, for n = 3 . second : you suppose, of course based on some expected outcome, but you just suppose that the thesis would be true for some value n = k . third : you try to prove mathematically , based on that supposition, that in that case the thesis will also be true for the next value n = (k + 1) . ((if not, then not, of course .)) conclusion: after the proof succeeded, and convinced , you can decide that based on the shown validity of the first value of n, i.e. n = 3, the thesis is also true for n = 4, and therefore for n = 5, because of this also for 6 and for each following natural number . all this based on the logical steps done from "second" to "third" , which provides a proof *Only* for the follower of the already *proven number* . in europe we call this kind of proof a proof by complete, or full, induction, inductio plena, vollständige Induktion, volledige inductie. good luck .

[@chengkaigoh5101]

@@keescanalfp5143 thank you

[@bhchoi8357]

😊

[@sarita9]

❤

[@sleeperzzz4931]

If this formula is known, it can indeed be proved. But how did the first person to figure out this formula do it?

[@keescanalfp5143]

think, suppose that when she was young like we were once, she tried on a piece of paper the first cubes 1 _ 8 _ 27 _ 64 _ 125 _ 216 .. then maybe tried their row of differences , 7 _ 19 _ 37 _ 61 _ 91 .. weird somehow , but yet isn't there a kind of regularity between these , 12 _ 18 _ 24 _ 30 .. well that is , there could be more than one interesting thing about or between or around them . then , on another day , trying their sums like in the video 1+8 , which produces 9 . and 1+8+27 , which produces 36 , hey nice. and 1+8+27+64 , producing 100 . uff, too nice, what's up here . what could be the buzz. all of them seem to be squares . squares of what ? what's between them ? etcetera .. triangle numbers ? how ? why ? and so on . all hypothetical , out of childish curiosity , nothing more .

[@sleeperzzz4931]

I asked chatgpt, and the first person to figure out the formula was Euler, a great mathematician.

[@keescanalfp5143]

​@@sleeperzzz4931, so beautiful that it was him ! does c.gpt know *How* the man came upon the formula or even upon the searching for it .

[@jotawski]

🌹🙏🌹

[@mahinchawla8705]

isnt this circular reasoning?

[@adw1z]

Not at all, just induction

[@RealLukifer]

The teacher first shows that the equation is true for n=1,n=2,n=3 Then he shows that it is always true for n+1/the next integer. So n=4 is true, n=5 is true, etc. equation is true by induction.

[@ToanPham-wr7xe]

😮

[@MGmirkin]

What about "the first n odd cubes"?

[@MGmirkin]

Just wondering from some stuff on Perfect Numbers & Mersenne Primes, which has it that a Perfect Number can be expressed as a sum of some number of consecutive odd cubes... e.g. 1^3+3^3+5^3+7^3, ... n^3 etc. Which I guess would basically just be (1+((1-1)*2))^3+(1+((2-1)*2))^3+(1+((3-1)*2))^3+(1+((4-1)*2))^3+...+(1+((n-1)*2))^3, where n is the ordinal of the odd cube (1st odd cube, 2nd odd cube, etc.)?

[@MGmirkin]

Would it just be the square of the sum of the [odd] numbers being cubed? Wolfram|Alpha seems to say "no." Not for cubes of first odd numbers. :\ Hmm...

[@amitavadasgupta6985]

{n(n+1)2}^2 is resultant ans

[@charl1878]

Is there an alternative proof without using induction?

[@tonybantu9427]

Yes. But you will need to know that there exist real numbers A, B, C,....SUCH THAT the sums of any given powers can be expressed in a closed form. As follows: 1^3 + 2^3 + 3^3+...+n^3 = A*n(n+1) + B*n(n+1)(n+2) + C*n(n+1)(n+3) = ( n^2.(n+1)^2 )/4 LHS: In this case, (A,B,C) = (1/2, -1, 1/4), hence the cubes sum to: ( n^2.(n+1)^2 )/4 But we also know that: 1+2+3+....+n = n(n+1)/2 RHS: Placing this value inside the right hand parenthesis gives: [ n(n+1)/2 ]^2 Which simplifies to: n^2.(n+1)^2 / 2^2 And is equal to the LHS.

[@chiragraju821]

@@tonybantu9427Why the strange choice of polynomials like n(n+1), n(n+1)(n+2)…? are they orthogonal or something?

[@hridayevyas8906]

So... you used the expression to prove the expression? I dont get how that proves the initial proof if you just claim the main question as the very first step of the solution?

[@PrimeNewtons]

That is called mathematical induction. You make a claim and then show your claim is true.

[@RoshanPaulThePhysicsShelter]

It can be done without induction

[@user-xq1fc6uy3d]

12:26 The face :))))

[@LeaderTerachad]

I don't know why but i thing you don't need to put thos intro in you're videos

[@Aenderson23]

I make de formula in another side of equation and get same result

[@caothai1770]

Giống quy nạp newton nhỉ

[@yangwang8038]

数学归纳法嘛

[@luizpereira1690]

Hummm

[@ahmeterturk6901]

Begendım kibar

[@user-qr7dw4hk6x]

Мат. Индукцией доказывается элементмрно

[@blackbolshevik]

数学归纳法秒解

[@alquinn8576]

the way you write a term and then go back to wrap in in parentheses (instead of making parentheses right away) is anxiety-inducing 😮‍💨

[@ibrahem_x564]

🇵🇸🇵🇸

[@user-lc8kz5bs8o]

@ ... better use [ abc ( def ) ghi ] ... @

[@holyshit922]

Prove that (cos(x))^{(n)} = cos(x+n*pi/2) where n means nth derivative

[@PrimeNewtons]

Is this from differential equations?

[@holyshit922]

@@PrimeNewtons No I used this while expanding exponential generating function of ChebyshovT polynomial E(x,t) = exp(xt)cos(sqrt(1-x^2)t) and i have seen lately this problem on one of the math forums In my opinion it is good exercise for mathematical induction if we are not allowed to use complex numbers

[@holyshit922]

I used Chebyshov not Chebyshev because your transcription of this name is poor and leads to misreading Name of this guy written in cyrylic has two dots over last e which is read as yo but it is simplified to o (Maybe because it would be difficult to read sh and yo but i dont know why this simplification occured)

[@PrimeNewtons]

I'm looking at this video soon

[@rajkumarbajagain5392]

Lets join you and me in a social media. I wanna share mathematical content with you. I am a permanent mathematics teacher of government of Nepal.

[@PrimeNewtons]

We can correspond by email. I also have Instagram @primenewtons.

[@blackcat71888]

준석아

[@oneli8492]

笨！变成积分两步就证明了🤣

[@user-jj8kg5ef2t]

There is a graphical solution. Unfortunately, cannot be shown on comments.

[@user-hi5fo3hb4b]

Оно доказывается иначе и гораздо быстрее через вывод общей суммы рядов

[@furikake_on_bread]

🌆🏙️🌃🌌🌉 🏙️🏙️🌃🌌🌉 🌃🌃🌃🌌🌉 🌌🌌🌌🌌🌉 🌉🌉🌉🌉🌉 Maybe able to prove visibly by using ↑ But I’m not sure how to explain cubed terms…

[@keescanalfp5143]

well your proposal seems to have a brilliant side . let's do a try . starting from the upper left side you could consider to see the unity, 1 = 1³ in a quarter around this, you could see 3 + 5 , that is twice the main value 4 , 8 = 2³. together with the starting unity you see lying in a square : 1+3+5 = 9 . in a quarter around these 9 you could lay down 7+9+11 squares , that is three times the main value of 9 , 27 = 3³ squares added, so together there's laying a larger square of 6×6 unity squares . in a quarter around these 36 you could add four hooked paths out of 13+15+17+19 unity squares, that is four times the main value of 16 , 32+32 = 64 added, so 4³ added to the already laying 1+8+27 ; the complete sum of squares is now 1+(3+5)+(7+9+11)+(13+15+17+19), laying in a square of 10×10 unities . in a quarter field around the starting unity . well we leave to your imagination how to continue with five hooked paths of unity squares around these 100 . good luck .