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
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.
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
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.
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 .
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
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 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.
@@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.
@@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 ?
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.
@@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.
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.
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)
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.
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.
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.
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.
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.
Powered by Chat Gpt: The goal is to prove that the expression I = (n/6) + (n^2)/2 + (n^3)/3 is an integer for all n in Z (integers). **Step 1: Use the Division Algorithm** For any integer n, there exist integers k and r such that: n = 6k + r where r ∈ {0, 1, 2, 3, 4, 5} Substitute n = 6k + r into the expression for I: I = (6k + r)/6 + ((6k + r)^2)/2 + ((6k + r)^3)/3 **Step 2: Expand the Powers of (6k + r)** (6k + r)^2 = 36k^2 + 12kr + r^2 (6k + r)^3 = 216k^3 + 108k^2r + 18kr^2 + r^3 Substitute these into the expression for I: I = (6k + r)/6 + (36k^2 + 12kr + r^2)/2 + (216k^3 + 108k^2r + 18kr^2 + r^3)/3 **Step 3: Simplify Each Term** Simplify each part of the expression: (6k + r)/6 = k + r/6 (36k^2 + 12kr + r^2)/2 = 18k^2 + 6kr + r^2/2 (216k^3 + 108k^2r + 18kr^2 + r^3)/3 = 72k^3 + 36k^2r + 6kr^2 + r^3/3 Now the expression for I becomes: I = k + 18k^2 + 72k^3 + 6kr + 36k^2r + 6kr^2 + (r/6 + r^2/2 + r^3/3) The terms involving k are integers, so we focus on the remaining part involving r: r/6 + r^2/2 + r^3/3 **Step 4: Check for Each r ∈ {0, 1, 2, 3, 4, 5}** Now we check whether (r/6 + r^2/2 + r^3/3) is an integer for each r: - For r = 0: r/6 + r^2/2 + r^3/3 = 0 (integer) - For r = 1: 6/6 (integer) - For r = 2: 30/6 (integer) - For r = 3: 84/6 (integer) - For r = 4: 180/6 (integer) - For r = 5: 330/6 (integer) **Conclusion:** In all cases, (r/6 + r^2/2 + r^3/3) is an integer for r ∈ {0, 1, 2, 3, 4, 5}. Therefore, I is an integer for all n in Z.
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.
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!
We can write the expression as ⅙ n(n+1)(2n+1). Now write (2n+1) as (n-1)+(n+2). Separate the expressions and you get ⅙ (n-1)n(n+1) + ⅙ n(n+1)(n+2). Now we can notice both the expressions are the product of three consecutive integers. We already know product of k consecutive integers is divisible by k!, bingo, both the expressions are divisible by 3! = 6 and so is their sum. Amazing video and work, sir!!
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
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?
Yes. Product of n consecutive integers is divisible by n!. Hence, the second step is enough to prove that both parts are divisible by 3! = 6. 3n(n+1) is divisible by 3 x 2 = 6 (n-1)n(n+1) is divisible by 3! = 6
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
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.
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
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!
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😅
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.
Easiest approach - Sum of 1²+2²+3²+....+n²= n(n+1)(2n+1)/6 =(2n³+3n²+n)/6 Thats our question, and since the sum of squares of integers from 1 will be an integer for any n, we can say that the given expression is am integer if n is an integer.
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}')
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.
Why not use modular arithmetics ? It is very easy to see that if n equals 0, or 1, or 2, or 3, or 4, or 5 modulo 6, then n (n+1) (2n+1) equals 0 modulo 6. Another story is to notice that n (n+1) (2n+1) / 6 is the sum of the n first square integers, then it is an integer (at least for n > 0). Thanks for your videos 🙂
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.
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.
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)
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.
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.
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
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
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
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.
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.
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.
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.
I used brute force haha So, if a number isn't divisible by 3 it can be represented as 3k + 1 or 3k + 2; which are 1 or 2 mod 3 respectively You can show that 1² = 1³ = 1 mod 3 and 2² = 1 mod 3, and 2³ = 2 mod 3 So what you're left to do is to show that 2n³ + 3n² + n in terms of k is 0 mod 3. So it's either 2 + 3 + 1 = 6 = 0 mod 3 or 4 + 3 + 2 = 9 = 0 mod 3 Obviously 0 + 0 + 0 = 0 mod 3 Thus it's always divisible by 3 no matter what n is. And divisibility by 2 is shown in terms of parity as presented in the video. Thus, it's also divisible by 6
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
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)
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
In fact, the product of the consecutive numbers is divisible by 6. Furthermore, 3 times two consecutive numbers is divisible by 6 two. So the proof is much shorter than what you did. Still cool and thanks for the content.
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.)
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
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.
Excellent! Can you make a video about how someone came up with the idea of creating this problem? I always wonder how someone mathematicians create math problems. Thanks
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.
I don't think this is a proof but just look at it. n/6+n^2/6+n^3/3=2n^3+3n^2+n/6= (2n+1)*(n+1)*n/6 and that formula is same as when we use a sigma for the square form. so integer's square is integer and integer and integer's addition is also integer so this the question is proofed. Is this right?
@@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
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.
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...
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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
Vvbbbbbhvv😅😅bbnbnygb😊
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.
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
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.
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.
Mulțumesc? Cu plăcere!@@SrisailamNavuluri
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 .
Yes. This was the approach I took as well. Simpler, faster.
But, I liked @primenewton's derivation as well.
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
Excellent explanation. Thank you for your generosity in sharing your knowledge, teacher.
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.
Not so simple! You have to show that as 6 divides n + 2n^2 + 3n^3 further steps that 6 evenly divides that integer.
Ohhps it as he was writing out 6 / ( n + 3n^2 + 2n^3 ) is another integer as he further explains!
@@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.
@@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.
@@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 ?
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.
Only for natural numbers. That's why I didn't use it.
"disability"? :D Surely you meant "divisibility"?
@@bjornfeuerbacher5514yes, edited now 💀
@@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.
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.
If you write n as 6*m+k where m is any integer and 0≤k
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)
👍... a very straightforward approach!!!
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.
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.
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.
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.
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.
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.
Cool. 😎
Powered by Chat Gpt:
The goal is to prove that the expression
I = (n/6) + (n^2)/2 + (n^3)/3
is an integer for all n in Z (integers).
**Step 1: Use the Division Algorithm**
For any integer n, there exist integers k and r such that:
n = 6k + r where r ∈ {0, 1, 2, 3, 4, 5}
Substitute n = 6k + r into the expression for I:
I = (6k + r)/6 + ((6k + r)^2)/2 + ((6k + r)^3)/3
**Step 2: Expand the Powers of (6k + r)**
(6k + r)^2 = 36k^2 + 12kr + r^2
(6k + r)^3 = 216k^3 + 108k^2r + 18kr^2 + r^3
Substitute these into the expression for I:
I = (6k + r)/6 + (36k^2 + 12kr + r^2)/2 + (216k^3 + 108k^2r + 18kr^2 + r^3)/3
**Step 3: Simplify Each Term**
Simplify each part of the expression:
(6k + r)/6 = k + r/6
(36k^2 + 12kr + r^2)/2 = 18k^2 + 6kr + r^2/2
(216k^3 + 108k^2r + 18kr^2 + r^3)/3 = 72k^3 + 36k^2r + 6kr^2 + r^3/3
Now the expression for I becomes:
I = k + 18k^2 + 72k^3 + 6kr + 36k^2r + 6kr^2 + (r/6 + r^2/2 + r^3/3)
The terms involving k are integers, so we focus on the remaining part involving r:
r/6 + r^2/2 + r^3/3
**Step 4: Check for Each r ∈ {0, 1, 2, 3, 4, 5}**
Now we check whether (r/6 + r^2/2 + r^3/3) is an integer for each r:
- For r = 0:
r/6 + r^2/2 + r^3/3 = 0 (integer)
- For r = 1: 6/6 (integer)
- For r = 2: 30/6 (integer)
- For r = 3: 84/6 (integer)
- For r = 4: 180/6 (integer)
- For r = 5: 330/6 (integer)
**Conclusion:**
In all cases, (r/6 + r^2/2 + r^3/3) is an integer for r ∈ {0, 1, 2, 3, 4, 5}. Therefore, I is an integer for all n in Z.
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.
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!
We can write the expression as ⅙ n(n+1)(2n+1). Now write (2n+1) as (n-1)+(n+2). Separate the expressions and you get ⅙ (n-1)n(n+1) + ⅙ n(n+1)(n+2). Now we can notice both the expressions are the product of three consecutive integers. We already know product of k consecutive integers is divisible by k!, bingo, both the expressions are divisible by 3! = 6 and so is their sum.
Amazing video and work, sir!!
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
1. Mathematical induction
2. Factorization and some conclusions
Yes indeed way you presented can be liked
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?
You are correct. Conclusion was incorrect.
Yes. Product of n consecutive integers is divisible by n!.
Hence, the second step is enough to prove that both parts are divisible by 3! = 6.
3n(n+1) is divisible by 3 x 2 = 6
(n-1)n(n+1) is divisible by 3! = 6
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
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.
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
=3n(n+1) + 2n(n²−1) both divisible by 6
can be recast as the formula for Σn²
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!
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😅
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.
Easiest approach -
Sum of 1²+2²+3²+....+n²=
n(n+1)(2n+1)/6
=(2n³+3n²+n)/6
Thats our question, and since the sum of squares of integers from 1 will be an integer for any n, we can say that the given expression is am integer if n is an integer.
this is just the sum of the squares of the first n positive integers, hence an integer. no divisibility proof needed.
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.
Wow! Such an inspirational video! It was pure joy to hear your explanations :)
Great video. You just messed up the conclusion. It's: "the expression" is an integer
AND NOT divisible by 6
😢
🙃🙃🙃
Looking through comments to find the one who pointed this out....
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}')
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.
If you simplify the expression, which you almost did gives, n.(n+1)(2n+1)/6, which is the formula for 1^2+2^2+3^2+...+n^2, which is an integer.
If n is an integracyjne, is enougt to show Thatcher only 2n^2 + 3n + 1 is div by 6 without rest
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.
Why not use modular arithmetics ? It is very easy to see that if n equals 0, or 1, or 2, or 3, or 4, or 5 modulo 6, then n (n+1) (2n+1) equals 0 modulo 6. Another story is to notice that n (n+1) (2n+1) / 6 is the sum of the n first square integers, then it is an integer (at least for n > 0). Thanks for your videos 🙂
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.
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.
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)
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.
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.
Notice 2n^3+3n^2+n=n(2n^2+3n+1)
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
I was just about to comment that😅
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
You can just prove it with math induction but it will involve a lot of calculations but not way more than in original method
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
Nice explanation! ❤❤
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
yep.. divisibility by 2 is immediately obvious. Then reduce it to the equivalent of looking at k(k+1)(k+2)/6 to confirm divisibility by 3.
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.
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.
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.
Exactly. Even easier if you plug in -1 instead of 2, mod 3.
n/6 + (n^2)/2 + (n^3)/3 = (n.(n+1).(2.n+1))/6 = (1^2) + (2^2) + (3^2) + .... + (n^2) is an integer because it is a sum of squares of integers...
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.
I used brute force haha
So, if a number isn't divisible by 3 it can be represented as 3k + 1 or 3k + 2; which are 1 or 2 mod 3 respectively
You can show that 1² = 1³ = 1 mod 3
and 2² = 1 mod 3, and 2³ = 2 mod 3
So what you're left to do is to show that 2n³ + 3n² + n in terms of k is 0 mod 3.
So it's either 2 + 3 + 1 = 6 = 0 mod 3 or
4 + 3 + 2 = 9 = 0 mod 3
Obviously 0 + 0 + 0 = 0 mod 3
Thus it's always divisible by 3 no matter what n is.
And divisibility by 2 is shown in terms of parity as presented in the video.
Thus, it's also divisible by 6
Problems like these can be easily solved with the knowledge of the below theorem :
"The product of N consecutive integers is divisible by N!"
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
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)
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
In fact, the product of the consecutive numbers is divisible by 6. Furthermore, 3 times two consecutive numbers is divisible by 6 two. So the proof is much shorter than what you did. Still cool and thanks for the content.
I proved by Induction Hypothesis and works like a glove
Can you prove that for a positive integer 'n', (n)(n+1)(2n+1) is always divisible by 6
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.)
Good question ❓😊
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
n + 3n^2 + 2n^3 = 6k
consider n mod 2 & n mod 3
I liked how you broke down the algebra.
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.
Isn't the sum of the square of the n first integer (1 + 2^2 + ... +n^2) ?
Therefore it must be an integer.
Excellent! Can you make a video about how someone came up with the idea of creating this problem? I always wonder how someone mathematicians create math problems. Thanks
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.
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
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
Good job thank you and keep it coming
Just prove when n=1, expression is integer. because any other number is just a multiplier.
Trivial using induction, isn't it?
Jejeje. Que bonita prueba. Muchas Gracias .
=(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
Mod 6 equivalency classes. 0:40
Is there any proof or disproof of this statement? The product of any n consecutive natural numbers is always divisible by n! (n factorial)
I don't think this is a proof but just look at it.
n/6+n^2/6+n^3/3=2n^3+3n^2+n/6= (2n+1)*(n+1)*n/6
and that formula is same as when we use a sigma for the square form.
so integer's square is integer and integer and integer's addition is also integer so this the question is proofed.
Is this right?
I liked this one a lot
proof my induction for the win
Let's get into the video
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
Factoring is incorrect
@@PrimeNewtons oh yea my bad, thx for your feedback!
@@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
Doesn't this show that "the product of n consecutive numbers is divisible by n!" ?
Bravo!
This is the sum of squares
Just test 0,1,2,3,4,5 mod 6 and it's over.
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.
But Prime Newtons method is more fun.
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.
Here's another which uses the same rule: is 6323(base 8) prime?
Interesting
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)
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...
Induction is very easy and you don't have to proof that each number is divisible by 2 and 3
6l (n +3n^2 +2n^3)
Check all 6 Casey qed
J'ai fait une récurrence, pour le plaisir.
Nicely done!