A fun proof for an integer

It is difficult to initially accept that this theorem is true until you prove it using basic algebra and number theory.

[@laitinlok1]

Use mathematical induction. Let P(n) =n/6 + n^2/2 + n^3/3. P(0)=0/6+0^2/2+0^3/3=0. P(0) is an integer. P(1)=1/6+1^2/6+1^3/6=1. P(1) is an integer. P(-1)= -1/6+ (-1)^2/2+(-1)^3/3=0. P(-1) is an integer. Assume P(k) is an integer, i.e. k/6+k^2/2+k^3/3=t, where t,k are integers. P(k+1)= (k+1)/6 + (k+1)^2/2 +(k+1)^3/6= (k+1)/6 + (k^2+2k+1)/2+(k^3+3k^2+3k+1)/3 = k/6 +k^2/2 +k^3/3 + 1/6 +(2k+1)/2 + (3k^2+3k+1)/3= t+2k/2 +(3k^2+3k)/3 +1/6+1/2+1/3= t+k+k^2+k+1 =t+k^2*+2k+1. P(k+1) is an integer. P(k-1) = (k-1)/6 +(k-1)^2/2+ (k-1)^3/3= (k-1)/6 +(k^2-2k+1)/2+ (k^3-3k^2+3k-1)/3= k/6 +k^2/2 + k^3/3 + (-2k+1)/2 + (-3k^2+3k-1)/3 = t-1/6+1/2-1/3-2k/2+(3k-3k^2)/3=t+0-k+k-k^2= t + k^2. P(k-1) is an integer. By mathematical induction, n/6 + n^2/2 + n^3/3 is an integer for all integer n.

[@SrisailamNavuluri]

If n,n+1 are not multiples of 3,n-1,n+2 are multumesc of 3 2n+1=n-1+n+2=multime if 3+ multiple of 3 That gives the answer.

[@viktor5534]

Mulțumesc? Cu plăcere!​@@SrisailamNavuluri

[@juanlopez-lp4lp]

asumimos que si n=m , el resultado es 'x' probamos con n=(m+1) , resultado 'y' restamos 'y'-'x' 'y'-'x' = n² + 2n + 1 = (n+1)² como n es integral , y-x es integral . si se cumple para n=0 , se cumple para n+1 . todo numero integral se puede definir como n+1 siendo n integral .

[@enerjae7174]

To check divisibility mod 3, we can add or subtract 3 to the (2n+1) term to make it a little easier n(n+1)(2n+1) (mod 3) ≡ 2(n-1)n(n+1) (mod 3) n(n+1)(2n+1) (mod 3) ≡ 2n(n+1)(n+2) (mod 3) In either case, we get a product of three consecutive integers so divisibility is clear

[@agytjax]

Problems like these can be easily solved with the knowledge of the below theorem : "The product of N consecutive integers is divisible by N!" . For e.g : n(n+1)(n+2) is always divisible by 3! = 6 Given expression can be written as : (n+ 3n^2 + 2n^3)/6 = n(1+3n+2n^2)/6 = n(2n+1)(n+1)/6 We can write 2n+1 = (n+2) + (n-1) So n(2n+1)(n+1)/6 = n (n+1) {(n+2) + (n-1)}/6 = {n(n+1)(n+2) + n(n+1)(n-1)}/6 = {n(n+1)(n+2)}/6 + {(n-1)n(n+1)}/6 The above expression is the sum of two 3-consecutive integers. Hence each of the sum is divisible by 6 and therefore the sum too, from the earlier theorem

[@adw1z]

nice, very similar to your previous divisibility video. Another fun fact / method is for n > 0: n/6 + n^2 /2 + n^3 /3 = 1/6 n(n+1)(2n+1) = 1^2 + 2^2 + … + n^2, which is always an integer clearly. For negative integers, let n > 0 and m = -n: we get m/6 + m^2 /2 + m^3 /3 = -n/6 + n^2 /2 - n^3 /3 = -(n/6 + n^2 /2 + n^3 /3) + n^2, which is always an integer since the first cubic term is from the positive case, and n^2 is.

[@PrimeNewtons]

Only for natural numbers. That's why I didn't use it.

[@bjornfeuerbacher5514]

"disability"? :D Surely you meant "divisibility"?

[@adw1z]

@@bjornfeuerbacher5514yes, edited now 💀

[@adw1z]

@@PrimeNewtonsif we let 0 < n -> -n, we get -n/6 + n^2 /2 - n^3 = -(n/6 + n^2 /2 + n^3 /3) + n^2, which is always an integer since the first cubic is from the positive case, and n^2 is.

[@jay_sensz]

If you write n as 6*m+k where m is any integer and 0≤k

[@I-am-what-you-call-useless]

This is the simplest problem. If you take 6 as LCM of Denominator and factorize the numerator thereafter you will end up with the expression of the sum of square of first n integers which is an integer of course.

[@lawrencejelsma8118]

Not so simple! You have to show that as 6 divides n + 2n^2 + 3n^3 further steps that 6 evenly divides that integer.

[@lawrencejelsma8118]

Ohhps it as he was writing out 6 / ( n + 3n^2 + 2n^3 ) is another integer as he further explains!

[@I-am-what-you-call-useless]

@@lawrencejelsma8118 Don't you know the expression of sum of squares of first n integers. The 6 in denominator is already accommodated in the expression. You don't need to show this.

[@lawrencejelsma8118]

@@I-am-what-you-call-useless ... No it wasn't!? Then why else did he have to prove further after getting the numerator n + 3n^2 + 2n^3 and to prove 6 divides into that evenly to produce another integer result. I don't believe your analysis in a quotient with 6 in the denominator forces a numerator must be a multiple of 6, say, 6m or proving n + 3n^2 + 2n^3 = 6m where m is another integer.

[@I-am-what-you-call-useless]

@@lawrencejelsma8118 By mathematical Induction as well as by calculation we can shpw that 1² + 2² + ... n² = ⅙n(n+1)(2n+1) Here LHS is an integer right ? So RHS must be an integer too. Now if we expand the RHS we will automatically get the expression in the question. = n/6 + n²/2 + n³/3 Do it yourself and check. Why can't you understand such simple logic ?

[@sovietwizard1620]

I've watched enough videos of yours to know how to solve problems like this without help and without even any knowledge on how to start. This means your videos are very very useful, thank you!

[@misterj.a91]

Great video. You just messed up the conclusion. It's: "the expression" is an integer AND NOT divisible by 6

[@ThenSaidHeUntoThem]

😢

[@PrimeNewtons]

🙃🙃🙃

[@robertcotton8481]

Looking through comments to find the one who pointed this out....

[@bjornfeuerbacher5514]

I would say the method shown by the professor for the divisibility by 3 is much too complicated. Simply use the factorization. There are three possibilities: 1) n is divisible by 3; done 2) n+1 is divisible by 3; done 3) neither n nor n + 1 is divisible by 3; only there we have a tiny bit of work to do. In that case, we can write n = 3k+1 and n+1 = 3k+2 with an integer k. But then 2n+1 = n + (n+1) = 6k + 3, which obviously is divisible by 3; done.

[@Grecks75]

For non-negative n, this is the formula for the n-th square pyramidal number, i.e. for the sum of the first n perfect squares. Therefore it must be an integer. For negative n, write n=-m for some positive m, and substitute in the expression. If you now add 6m/6 (an integer!) to it, you can reduce it to the first case, getting an integer. So, also for negative n, the expression always yields an integer. Edit: Added an argument for the negative n case.

[@ForestHills101]

Another approach: = 1/6*n(n+1)(2n+1) =1/24*(2n)(2n+1)(2n+2) The first and last terms are divisible by 2 and 4 and one term by 3. Therefore the product is divisible by 24

[@Grecks75]

Another simple way to show that n(n+1)(2n+1) is divisible by 3: 3n^2(n+1) is clearly divisible by 3, as is (n-1)n(n+1) as a product of 3 consecutive integers. If you subtract the latter from the former you get just n(n+1)(2n+1) which therefore must also be divisible by 3.

[@alexlacerdaleite1304]

Excellent explanation. Thank you for your generosity in sharing your knowledge, teacher.

[@skwbusaidi]

The last eexpresion , we can directly say that the expresion is divible by 6 3n^2(n+1).-(n-1)n(n+1) The first part at n or n+1 is even which multiply by 3 that make it divisible by 6. The second part are product of three consecutive number which always is divisible by 6

[@godQlol]

n( 1/6 + n/2 + n^2 /3) = n(1 + 3n + 2n^2)/6 = n(n + 1)(2n + 1)/6 where n(n + 1)(2n + 1) is a multiple of 2 and multiple of 3 and therefore, a multiple of 6. QED Prove for multiple of 2: It's obvious that n(n + 1) is a multiple of 2 which implies n(n + 1)(2n + 1) is also a multiple of 2 Prove for multiple of 3: It's trivial to see that n = 3k implies n(n + 1)(2n + 1) is a multiple of 3 We use mod 3 in the second case, n(n + 1)(2n + 1) ≡ n(1 + n)(1 - n) (mod 3) => n(1 - n^2) It's easy to notice that n^2 ≡ 1 (mod 3) when n isn't a multiple of 3, therefore n(n + 1)(2n + 1) is a multiple of 3

[@Mrcasgoldfinch]

In the case of divisibility by 3, you can complete the remaining possibilities: n = 3k + 1 (then n(n+1)(2n+1) = (3k+1)(3k+2)(6k+3) - the last term is divisible by 3) and n = 3k + 2 - then the product is (3k+2)(3k+3)(6k+5), the middle term is divisible by 3.

[@godQlol]

i would say that its better if u transform n(n + 1)(2n + 1) into n(1 - n)(2 - n) which u can easily notice that for any integer, there will always be a factor turning it into a multiple of 3

[@Metaverse-d9f]

the original expression =n(n+1)(2n+1)/6=1*1+2*2+3*3+...+n*n, the sum of integers is an integer, so the original expression is also an integer.

[@davidplanet3919]

The numerator can be written 3n(n+1) + 2n(n-1)(n+1). Both terms are divisible by 2 and 3. The first term has two consecutive numbers. The second has three consecutive numbers.

[@user-jl7sd8gm6l]

A simpler way to solve this is to write n/6=(n/2)-(n/3) and just pair up the two with same denominator and in resulting numerator you will see integral multiple of denominator.

[@Grecks75]

Cool. 😎

[@electricportals3644]

A while back I derived the formula for the sum of the squares of the integers 1 to n, and when I saw the thumbnail I recognized it as that formula! Building from there, all squares of integers are also integers and the sum of integers also result in an integer.

[@yulia6354]

Wow! Such an inspirational video! It was pure joy to hear your explanations :)

[@scottparkins1634]

The way I approached this was let the value be given by f(n), then a direct calculation gives f(n+1) - f(n) = (n+1)^2. Hence since f(1) = 1 (natural) by mathematical induction f(n) is always a natural number This approach also points us towards the fact that f(n) is the sum of the squares of the first n natural numbers.

[@agytjax]

Problems like these can be easily solved with the knowledge of the below theorem : "The product of N consecutive integers is divisible by N!"

[@TaiserBinJafor]

Very simple. If n is an integer, then: 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6 = (2n^3 + 3n^2 + n)/6 = n/6 + n^2/2 + n^3/3 Since the left side is the sum of the squares of all positive integers from 1 to n, it's definitely an integer. So, the right side must be an integer. Therefore, n/6 + n^2/2 + n^3/3 is also an integer. (Showed)

[@okohsamuel314]

👍... a very straightforward approach!!!

[@davidseed2939]

=3n(n+1) + 2n(n²−1) both divisible by 6 can be recast as the formula for Σn²

[@user-ji5su2uq9m]

n/6 + n^2/2 + n^3/3 = n(2n^2 + 3n + 1)/6 = n(n + 1)(2n + 1)/6 = (n - 1)n(n + 1)/3 + n(n + 1)/2 --(1) since product of 3 consecutive integers = multiples of 3, product of 2 consecutive integers = multiples of 2 (1) should be integer.

[@RyanLewis-Johnson-wq6xs]

Notice 2n^3+3n^2+n=n(2n^2+3n+1)

[@pseudo_goose]

Haven't watched yet, but I had a vibe that it was somewhat related to n(n+1)/2 being an integer, and I was right! This is just the expanded form of n(n+1)(2n+1)/6 (which famously is the sum of the first n squares), and it's fairly clear to see that the factors in the numerator always contain a 2 and 3 to make it a multiple of 6.

[@eduardoyamakawa1754]

You just forgot to finalize your conclusion that it is indeed an integer. Since n is an integer then 2n^3+3n^2+ n is also an integer and since this expression is also divisible by 6 as shown above then by definition of divides/divisible there exist K such that 6k= 2n^3+3n^2 + n Therefore n/6 + (n^2)/2 + (n^3)/3 is an integer

[@9ybruhwt491]

brilliant! I did... the proof that it is divisible by 3 uses the fact that one of the three consecutive numbers, n, n+1, or n+2, is divisible by 3. If either n or n+1 is divisible by 3, then the term n(n+1) is divisible by 3. If n+2 is divisible by 3, then 2n+1 = 2(n+2) - 3 is divisible by 3. Therefore, in any case, n(n+1)(2n+1) is divisible by 3.

[@shasha22in]

I used a similar approach but took the single fraction to be factorised as n(2n+1)(n+1) and proved that the product always divides 3 for all n ≡ 1 and 2 (mod 3)

[@RexxSchneider]

At 6:46 you ought to be saying that you are writing 2n^3 as 3n^3 - n^3 so that you can create a part of the expression that is divisible by 3. It helps students understand the motivation for that. Optionally you can continue with showing that n(n+1)(2n+1) is divisible by 3 like this: If 3 | n or 3 | (n+1) then you have your proof, but if neither is true, then it must be true that 3 | (n+2). In that case, set n+2 = 3m (where m is an integer). That means that n = 3m-2 and therefore (2n+1) = 6m-4 + 1 = 6m - 3, which is clearly divisible by 3. There's your proof. Recognising that 3 must divide one of n, (n+1), or (n+2) is the key to solving most of these sort of problems.

[@itsphoenixingtime]

If you make it equal to 1 fraction you actually get n(2n+1)(n+1)/6, which is the general sum for n squares from 1 to n. Since squares of integers are integers and adding integers to integers gives integers then you will have an integer. Hence it is only fair the expression is an integer, being the generalised form of the sum of squares from 1 to n. 1^2 + 2^2 + .... + n^2 = n(n+1)(2n+1)/6 = n(2n^2 + 3n + 1)/6 = 2n^3 + 3n^2 + n / 6 = n^3/3 + n^2/2 + n/6

[@alessandro_can3792]

fun fact, n(n+1)(2n+1)/6 not only is always an integer but it's the sum of the first n square numbers

[@FirstNameLastName-mw1pj]

If n is even, (n+3n^2)mod6 will always be the same as nmod6. If n is odd, it'll be (n+3)mod6. Thus adding the first 2 terms together will always give a non-integer component of 0 for nmod3=0, 2/6, which simplifies to 1/3 for nmod3=2, and 4/6=2/3 for nmod3=1. For nmod3=0, (0mod3)^3=0mod3, so the n^3 term is an integer. For nmod3=1, 1mod3*1mod3=1mod3*0mod3+1mod3=0mod3+1mod3=1mod3, thus the n^3 term would wind up with a fractional part of 1/3, which adds with the 2/3 previously obtained to get an integer. Likewise 2mod3*2mod3=1mod3, and 2mod3*1mod3=2mod3, so (2mod3)^3 is 2mod3, giving a fractional part of 2/3, which adds to the 1/3 previously obtained to give an integer.

[@doctorno1626]

Can you prove that for a positive integer 'n', (n)(n+1)(2n+1) is always divisible by 6

[@nasrullahhusnan2289]

⅙n+½n+⅓n²=⅙n(1+3n+2n²) =⅙n(n+1)(2n+1) Have to show that 6|n(n+1)(2n+1) Note that n and n+1 are two consecutive integers, hence n(n+1) and is even. 2n+1 is the sum of the n and n+1. Suppose n is not divisible by 3. Thus mod(n,3) is either 1 or -1 If mod(n,3)=-1 then 3 | n+1. Hence 3 | ⅙n+½n²+⅓n³. If mod(n,3)=1, mod[(n+1),3]=2 then mod[(2n+1),3] =mod(n,3)+mod[(n+1),3] =1+(-1) =0 --> 3 | n(n+1)(2n+1) As n(n+1)(2n+1) is an even integer and divisible 3 --> 6 | n(n+1)(2n+1) meaning that⅙n+½n²+⅓n³ is an integer.

[@bearantarctic5843]

Another solution: Show that (2n^3+3n^2+n)/6 is the sum of the squares of integers from 1 to n Because integers are closed under addition and multiplication, this sum is also an integer

[@kaenemorerinyane9392]

I was just about to comment that😅

[@fastneuro9829]

You can just prove it with math induction but it will involve a lot of calculations but not way more than in original method

[@BartBuzz]

If you let x= n+3n^2+2n^3, then when n=1 => x=6. Therefore for any n, x=6 times an integer. That tells me that the original expression is always an integer for any real number, n.

[@adampiechuta5774]

n(n+1)(2n+1) is divided by 2 because n or n+1 is diveded by 2 and divided by 3 because if n is divided by 3 (0mod3) or n+1 is divided by 3 so n=3k+2 (2mod3) or if n=3k+1 (1mod3) so 2n+1 is 2(3k+1)+1 is 6k +3 is divided by 3 so all product must be divided by 6

[@w.nickel2792]

It would be nice to put problems like this into a broader context. It is not unusual, that polynomials with rational coefficients produce integer values for integer arguments. For example, the binomial coefficients (n choose k) produce polynomials of that kind for fixed values of k.

[@nanamacapagal8342]

ATTEMPT: Combine the fractions. n/6 + (n^2)/2 + (n^3)/3 = (n + 3n^2 + 2n^3)/6 Factor = n(1+n)(1+2n)/6 For n = 0, the expression is also 0, an integer. For positive n, This is the formula for the sum of squares up to n. = 1^2 + 2^2 + ... + n^2 Which is clearly an integer. For negative n, this is the sum of squares from (-n+1) to 0. = (-n+1)^2 + (-n+2)^2 + (-n+3)^2 + ... + (-1)^2 + 0^2 Which is obviously an integer. Proof complete

[@MichaelBizaare]

Saw this in Tom M. Apostol's Calculus -- It's the sum of n integers squared: [1^2 + 2^2 + 3^2... +n^2] It's so weird... I've proven something to myself with this video. Thank you!

[@antoinegrassi3796]

Un joli petit exercice qui pousse à la réflexion, avec une présentation claire et propre, faite d'une voix calme et paisible. Merci, car ceux qui débitent leur texte comme une mitraillette, qui ecrivent comme des cochons et effacent leur texte aussitôt ecrit, sont insupportables. 1) Une petite récurrence sur 2n³+3n²+n, marche très bien. 2) On a aussi le remplacement de 2n³ par 3n³-n³ ce qui donne 3n³+3n²-n³+ n = 3n²(n+1) - n(n-1)(n+1), on obtient deux termes. Il suffit de "bien" les lire pour voir que chacun d'eux est un multiple de 6. En effet 3n²(n+1) est multiple de 3 (il y est en clair) et de 2 grâce aux deux facteurs consecutifs n(n+1). Et (n-1) n(n+1) produit de 3 facteurs consécutifs est multiple de 2 et 3 donc de 6 aussi. La différence des deux termes et donc aussi un multiple de 6. 3) mais le plus beau, c'est quand même que même (2n³+3n²+n)/6 = n (n+1)(2n+1)/6 qui est exactement la somme des n premier carrés (oui, oui) qui est bien un nombre entier. Lol😅

[@holyshit922]

1. Mathematical induction 2. Factorization and some conclusions Yes indeed way you presented can be liked

[@SALogics]

Nice explanation! ❤❤

[@ChristopherBitti]

This is the same as showing n + 3n^2 + 2n^3 is divisible by 6, which is the same as showing it's divisible by 2 and 3. Using the fact that n^p = n (mod p) for all primes p and integers n, this is trivial. n + 3n^2 + 2n^3 = n + 3n = 4n = 0 (mod 2) n + 3n^2 + 2n^3 = n + 2n = 3n = 0 (mod 3)

[@prawnydagrate]

Am I missing something or is the conclusion at the end wrong? Isn't it 2n³ + 3n² + n that's always divisible by 6, and the original equation which is supposed to be just any integer?

[@PrimeNewtons]

You are correct. Conclusion was incorrect.

[@nothingbutmathproofs7150]

Nicely done!

[@Dr_piFrog]

Combine fractions, then the numerator always produces a number divisible by 6. Now if you want a Python script to verify this, then look below. """ (n/6) + ((n**2)/2) + ((n**3/3) Can be transformed into: numerator = n + 3*(n**2) + 2*(n**3) denominator = 6 """ def top(value) -> int: value = value + 3*(value**2) + 2*(value**3) return value b = 6 for j in range(1,100): a = top(j) div = a/b print(f'Top = {a} and Bottom = {b}, thus the integer {div}')

[@kamaljain5228]

this is just the sum of the squares of the first n positive integers, hence an integer. no divisibility proof needed.

[@PrimeNewtons]

You mean first n natural numbers. Zero and negative integers not included. There's nothing like 'first n integers'. That's why I didn't use it.

[@gdtargetvn2418]

Both derivation already proves it is divisible by 6. First derivation: P = 2n^3+3n^2+n = n(n+1)(2n+1) is indeed divisible by 6 if: - You notice this being a part of the infamous identity: 1^2+2^2+3^2+...+n^2 = n(n+1)(2n+1)/6 (this is the best derivation that ends the problem rightaway, ONLY if n is a natural number, which, in this case, it's not). - Alternatively, you can check for n ≡ 0, 1, 2 (mod 3), which all should result in P divisible by 3. Second derivation is much easier to notice: 2n^3+3n^2+n = 3n^2(n+1) - (n-1)n(n+1) is indeed divisible by 6 because - n(n+1) is divisible by 2, so 3n^2(n+1) is divisible by 6. - (n-1)n(n+1) is divisible by 6.

[@namanhnguyen7933]

call A = n/6 + n^2/2 + n^3/3 notice that: n^2-n=0 mod 2; n^3-n=0 mod 3 (small Fermat theorem) that means (n^2-n)/2 and (n^3-n)/3 are integers ---> (n^2/2 - n/2) + (n^3/3 - n/3) + (n/6 - n/6) = (n^2/2 + n^3/3 + n/6) - (n/2 + n/3 + n/6) = A - n is an integer. n is an integer ---> A is also an integer

[@antonionavarro1000]

Excellent development of the exercise. Elegant, simple and straight to the core. But I would like to see a proof of the theorem that the product of three consecutive numbers is always a multiple of 3.

[@simplebutpowerful]

Let n be an integer. Then one of the following is true: n = 3k (k being an integer), hence n * (n+1) * (n+2) is an integer. n = 3k+1, then n+2 = 3k+3, hence n * (n+1) * (n+2) is an integer. n = 3k+2, then n+1 = 3k+3, hence n * (n+1) * (n+2) is an integer. In summary: one of the three consecutive integers is a multiple of three, which guarantees the product will be divisible by 3.

[@carly09et]

n + 3n^2 + 2n^3 = 6k consider n mod 2 & n mod 3 I liked how you broke down the algebra.

[@kragiharp]

How do you come up with all these problems to solve? (Relatively complicated expression - often with some kind of symmetry - turns out to be simple or have a simple property.)

[@eduardoyamakawa1754]

I proved by Induction Hypothesis and works like a glove

[@user-wx8bx4tu8w]

Good job thank you and keep it coming

[@benardolivier6624]

I don't understand why using such a complex approach to prove a simple fact. All you have to do is recognize that n/6 + n²/2 + n^3/3 is equal to n(n+1)(2n+1)/6 which is the formula for the sum of the n first squares. That is clearly an integer...

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@robertlockwood3425]

This might be a duplicate. Let k = n/6 + (n^2)/2 + {n^3)/3; then (n + 3n^2 + 2n^3)/6 = k; then n + 3n^2 + 2n^3 = 6k = 6k. but the left hand side is an integer so 6k is an integer so k is an integer And the expression mod 6 is zero.

[@gvssen]

This comment is slightly besides the point. I trust that for positive n, (n/6)+[(n^2)/2]+[(n^3)/3] = (1/6).n.(n+1).(2n+1) happens to be the sum of squares of first n positive integers as well. Hence for positive n the the concerned sum has to be an integer it self.

[@josephassy4225]

Well n(n+1)(2n+1)/6 is actually the summ of n^2 and if n is an integer the sum of n^2 will also be

[@seyda4184]

Good question ❓😊

[@karthikeyank132010]

Is there any proof or disproof of this statement? The product of any n consecutive natural numbers is always divisible by n! (n factorial)

[@kappasphere]

Let k=n/6+n²/2+n³/3. Therefore, n+3n²+2n³=6k. This is equivalent to m:=n+3n²+2n³=0 mod 6. For this, show that m=0 mod 2 and m=0 mod 3. 0+3(0)²+2(0)³=0 (mod 2 or mod 3) 1+3(1)²+2(1)³=6=0 (mod 2 or mod 3) 2+3(2)²+2(2)³=30 mod 3 This finishes the proof.

[@dalex641]

n(n+1)(2n+1) => plug n+1 instead of n => (n+1)(n+2)(2n+3). If n%3 = 0 => (2n+3)%3 =0 If n%3 = 1 => (n+2)%3 =0 if n%3 = 2 => (n+1)%3 =0

[@decidueye98]

Doesn't this show that "the product of n consecutive numbers is divisible by n!" ?

[@lucmacot5496]

Bravo!

[@MrCarlosmario22]

Jejeje. Que bonita prueba. Muchas Gracias .

[@chengshengway]

Just prove when n=1, expression is integer. because any other number is just a multiplier.

[@giuseppemalaguti435]

=(1/6)n(1+3n+2n^2)=(1/6)n*2(n+1)(n+1/2)=(1/6)n(n+1)(2n+1)=sommatoria dei quadrati dei primi n naturali

[@stottpie]

Let's get into the video

[@pyrite2060]

proof my induction for the win

[@fekundulo]

Trivial using induction, isn't it?

[@9adam4]

Mod 6 equivalency classes. 0:40

[@vasilisr7]

This is the sum of squares

[@EMelo-gs4sr]

Just test 0,1,2,3,4,5 mod 6 and it's over.

[@810fox]

Induction is very easy and you don't have to proof that each number is divisible by 2 and 3

[@davidzanetti1269]

Here's a prob. Is 111111 (base five) odd or even? How about 387631 (base nine)? If you know the trick you see at a glance that they are both even.

[@davidzanetti1269]

Here's another which uses the same rule: is 6323(base 8) prime?

[@BRSCute]

n + 3n^2 + 2n^3 = n( n + 1 ) + 2n( n + 1 ) = n ( n +2 ) ( n + 1 ). Sum of 3 consecutive numbers divisible by 6 => Proof done

[@PrimeNewtons]

Factoring is incorrect

[@BRSCute]

@@PrimeNewtons oh yea my bad, thx for your feedback!

[@BRSCute]

@@PrimeNewtons i fixed it: ( 2n + 1 ) ( n + 1 ) n. n( n + 1 ) divisible by 2. If n ≡ 1 ( mod 3 ): 2n + 1 ≡ 0 ( mod 3 ), making ( 2n + 1 ) ( n + 1 ) n divisible by 6 If n ≡ 2 ( mod 3 ): n + 1 divisible by 3, making ( 2n + 1 ) ( n + 1 ) n divisible by 6. If n ≡ 0 ( mod 3 ): ( 2n + 1 ) ( n + 1 ) n divisible by 6. I hope its correct now

[@chouchfroukh]

Interesting

[@konraddapper7764]

6l (n +3n^2 +2n^3) Check all 6 Casey qed

[@Specters1]

Cuz 2|(n)(n+1) if 6 | (n)(n+1)(2n+1) then 3| n(n+1)(2n+1) 1.) n = 3k ; true 2.) n+1 = 3k ; true 3.) n+2 = 3k n = 3k-2 from n(n+1)(2n+1) = n(n+1)(6k-4+1) = n(n+1)(6k-3) = 3 * n(n+1)(2k-1) and 3 | 3A so 6 | (n)(n+1)(2n+1)

[@KarlFredrik]

Easiest is to just recognize that the expression is equal to the sum of squares from 1 to N. Sum[k^2] for k=1 to N. Since this is trivially integer the expression is integer also.

[@KarlFredrik]

But Prime Newtons method is more fun.

[@ensiehsafary7633]

Bro this is just the formula for sum of n^2 😂 I proves it a few years ago

[@RyanLewis-Johnson-wq6xs]

Proof:n/6+(n^2)/2+(n^3)/3=(2n^3+3n^2+n)/6

[@robertveith6383]

That is not a proof. It is a step.

[@user-sm2ku3uq9y]

The best way to show divisibility is to use modular arithmetic.

[@robertveith6383]

No, it is not the best way.

[@user-sm2ku3uq9y]

@@robertveith6383 it Always works.

[@joergholzhauer3218]

Sorry, but you want to proof, that n/6+n²/2+n³/3 is an integer. NOT: "Its divisible by 6" (which is not true, if you take for example n=2, resulting in 5) 😊

[@robertveith6383]

No, he said if the "top (numerator) is divisible by 6." So, in your case where n = 2, he would agree that 18 is divisible by 6.

[@vasilisr7]

This is the sum of squares

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student

[@xiamojq621]

Please can you upload videos on probability for quant and data science student