Because n! ≥ n*(n-1)*(n-2) and n*(n+1)/2 < n*(n+1) for all n≥4, all we need to show is that (n-1)*(n-2) > n+1 n² - 3n + 2 > n + 1 n² + 1 > 4*n n + 1/n > 4, which is true for n ≥ 4 since 1/n > 0
Excellent lecture ! Induction works smartly, with ± heavy writings… Let’s try directly : in order to prove that n ! > n (n + 1) / 2 for any n ≥ 4, equivalent to 2 (n-1) ! > n + 1, we notice that (for n ≥ 4 ) : 2 (n-1) ! > 2^(n-1) because 2 (n-1) ! = 2x2x3x…x(n-1), and 2^(n-1) > n + 1 by studying f(x) = 2^x - x - 2 for x ≥ 3 (easy stuff…). Then we’ve got it… Thank you for your interesting videos ! 🙂
An other way is : let U_n = 2 (n-1) ! and V_n = n +1. Then U_n+1 / U_n = 2 n and V_n+1 / V_n = (n+2) / (n+1). As U_3 = V_3 = 4, and for n ≥ 1, U_n+1 / U_n > V_n+1 / V_n, Then V_n < U_n for n ≥ 4.
Nice presentation! Note, however, that it’s not necessary to use math induction. Claim: If n>3, then n! >(n(n+1))/2. Multiplying both sides by 2 and dividing by n, we get the equivalent inequality 2(n-1)!>n+1. As 2(n-1)!>2(n-1), to prove the claim it will suffice to show that 2(n-1)>n+1, for n>3. But this is immediate: 2(n-1)>n+1 iff 2n-2>n+1 iff 2n-n>2+1 iff n>3.
@Salko_ So to get (n+l)! > (n+1)(n+2)/2, he took n! > n(n+1)/2 and replaced n with n+ 1? And then to get (n+1)n! > (n+1)n(n+1)/2, he took n! > n(n+1)/2 and multiplied both sides by n+1?
@Myhair0_0 That's why we have to check whether the inequality holds for the base case (n=4 in this example) Then, the proof from the video, which @Salko_ summarized, shows that if the inequality holds for any integer >=4, then it holds for the next integer. Thus, after manually verifying that the inequality holds for n=4 and completing the short proof, we know that the inequality holds for n=4, n=4+1=5, n=5+1=6, and so on. Hope this helps
@@Salko_ I don't know why the freak my response got deleted last night, I was asking this: Are you saying (n+1)! > (n+1)(n+2)/2 come from replacing n with n+1 in the original n! > n(n+1)/2? And does (n+1)n! > (n+1)n(n+1)/2 cone from the original but after multiplying both sides by n+1?
Hello sir, Very interesting approach to the problem. But if we are looking for the condition of the equality happening is it not easier? What needs to happen so n*(n+1)/2=n! ? Since n!=0 we can divide by n so (n+1)/2=(n-1)! n+1=2(n-1)! n=2(n-1)!-1 If we replace n by k+1 to have a nicer number inside the factorial k=2k!-2 so k=2(k!-1) which means that k>k!-1 or k+1>k! and we can easily verify that this proposition is only true for very small values
Very rough, not rigorous idea but... I figured that because the factorial grows much faster than the quadratic, after some point there won't be any more solutions, so there isn't any need to check every case. I remembered the 1 + 2 + 3 = 1 x 2 x 3 meme, so n = 3 n also = 1. That was basically my reasoning, because after n = 3 the factorial grows much faster than the quadratic so they will never ever intersect again, the only need is to check for answers within that range. I think the proof of the factorial outlasting the triangular numbers for n >= 4 was rigorous to help cement that idea that they can never be equal and hence no solutions for that region.
@@ishanpurkait9124 It's a VERY difficult book. I'll recommend that you start with basic books about calculus, elementary number theory, classical plane geometry, linear algebra, and abstract algebra.
I immediately know of 1, and 3. I think those are the only 2 but proving that is a whole different issue. I might be able to do that but not sure exactly what I'd do maybe induction or contradiction but would take awhile to prove it.
Something bugged me, in the induction, since we only needed n>=2 why couldn't we start at 2 ? well, 15 years after formerly learning about induction, I finaly understand how important the initial step is (base case). if we start at 2, initial step would be 2! = 2 2+1 = 3 2 is not greater than 3 So we can't start with 2, And we can't start with 3 either since we have shown it is equal. This is a marvelous Induction.
There are exactly 77000 ordered quadruples (a,b, c,d) such that gcd (a,b, c,d) =77 and lcm (a,b, c,d) =n, What is the smallest possible value of n? Hello teacher. Could we look at this question? There were many solutions that i didn't understand well. I would like to see your approach
From the thumbnail: the solutions are n=1 and n=3 . For any n>3, the cumulative product is greater than and also will be increasing faster than the cumulative sum. I'm watching the video to see if you consider n=0 a solution or not (and why).
I thin, the o natural number satisfying this equation are n=1 and n=3. The left side can be substituted b n*(n+1)/2 (according to gauss forula for sumof the first n natural numbers) while the rigtside is equall to n! (accordingto the definition of factorial).So we seach values for n with n*(n+1)/2=n!. Since n! is per definition n*(n-1)!, we can transform this equation to (n+1)/2=(n-1)!. n=1 and n=3 are possiblesolutionsforthhis equation, because (1+1)/2=(1-1)! and (3+1)/2=(3-1)!. Equal numbers can not fullfill the equation, because the right side is alwas a natural number, while the left side is for even values of n neer an integer. For all odd numbers n greater than 3, (n-1)! is greater than (n+1)/2, so n=1 and n=3 are the only solutions. For me, it was obvious,that (n-1)! is greater than (n+1)/2 for an n>3, but nice, that you gave a proof ...
For a triangular number to be equal to a factorial, we get: n(n+1)/2 = n! so n+1 = 2(n-1)! thus n is odd. Let n = 2k+1, then k+1 = (2k)! and since 2k(2k-1) ≤ k+1 means (k-1)(4k+1) ≤ 0, so k ≤ 1, we have 2k(2k-1) > k+1 for all k > 1. This concludes: (2k)! ≥ 2k(2k-1) > k+1 for k > 1, therefore k ≤ 1, leaving k = 0 and k = 1 the only options. These both lead to valid solutions in n: n = 1 and n = 3.
ATTEMPT: By inspection, N = 1, 3 1 = 1 1 + 2 + 3 = 1 * 2 * 3 = 6 N = 2 doesn't work. 1 + 2 = 3, 1 * 2 = 2 For N >= 4: N! > N(N-1) = N^2 - N = (N^2)/2 + N/2 + (N^2)/2 - 3N/2 >= (N^2)/2 + N/2 + 8 - 6 > (N^2)/2 + N/2 = N(N+1)/2 Which is the sum of all natural numbers up to N. Therefore for all natural N >= 4, 1 + 2 + 3 + ... + N < 1 * 2 * 3 * ... * N, and so the two sides cannot be equal. The only solutions are N = 1 and N = 3.
The one slight issue is the substitution of n = 4 in the (n²-3n)/2. We would just need to show that (n²-3n)/2 is bigger than 0 for n ≥ 4. But that's easy.
Me: "Obviously its n = 1 or n = 3" *Trying to prove that these are the ONLY values* Me: You got me there LMAO I thought about using induction to prove it since I saw that n = 4 didn't work and I knew for sure n > 4 didn't work either but I was a bit apprehensive knowing that it would've required some work.
@@robertveith6383 you are right, yes n=1 xor n=3 As a fact I can write you that in my language there is a word "albo" which suits the best here and it is equivalent to exclusive or which you don't use
For a, b, c, d, e ∈N If a + b + c + d + e = abcde Find the maximum possible value of max {a, b, c, d, e } Sir pls explain it You've explained it before but there are different solutions at different places and the answer got by u also don't satisfy it
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
You are the best mathematics educator I've ever seen on RUclips. You neither overexplain nor underexplain. Every step is 100% clear.
Because n! ≥ n*(n-1)*(n-2) and n*(n+1)/2 < n*(n+1) for all n≥4, all we need to show is that (n-1)*(n-2) > n+1
n² - 3n + 2 > n + 1
n² + 1 > 4*n
n + 1/n > 4, which is true for n ≥ 4 since 1/n > 0
Never stop learning, nice lecture, you are genious.
Excellent lecture !
Induction works smartly, with ± heavy writings…
Let’s try directly : in order to prove that n ! > n (n + 1) / 2 for any n ≥ 4, equivalent to 2 (n-1) ! > n + 1, we notice that (for n ≥ 4 ) :
2 (n-1) ! > 2^(n-1) because 2 (n-1) ! = 2x2x3x…x(n-1), and
2^(n-1) > n + 1 by studying f(x) = 2^x - x - 2 for x ≥ 3 (easy stuff…).
Then we’ve got it… Thank you for your interesting videos ! 🙂
An other way is : let U_n = 2 (n-1) ! and V_n = n +1.
Then U_n+1 / U_n = 2 n and V_n+1 / V_n = (n+2) / (n+1).
As U_3 = V_3 = 4, and for n ≥ 1, U_n+1 / U_n > V_n+1 / V_n,
Then V_n < U_n for n ≥ 4.
I am thankful for your efforts, your videos are always nice and filled with benefits, it's our pleasure to watch your videos.
Nice presentation! Note, however, that it’s not necessary to use math induction.
Claim: If n>3, then n! >(n(n+1))/2. Multiplying both sides by 2 and dividing by n, we get the equivalent inequality 2(n-1)!>n+1. As 2(n-1)!>2(n-1), to prove the claim it will suffice to show that 2(n-1)>n+1, for n>3. But this is immediate: 2(n-1)>n+1 iff 2n-2>n+1 iff 2n-n>2+1 iff n>3.
I realized that while watching the video. Thanks.
It's really interesting sir make more videos like this sir 💯💯💯❣️💫✨🌟
Excellent job
Great video. While comparing n^2 + n with n + 2, you could also see that since n >= 4, n^2 >= 16 so n^2 + n >= n + 16 > n + 2
Nice point
9:42 where did that new n+1 on the RHS come from?
@Salko_ So to get (n+l)! > (n+1)(n+2)/2, he took n! > n(n+1)/2 and replaced n with n+ 1? And then to get (n+1)n! > (n+1)n(n+1)/2, he took n! > n(n+1)/2 and multiplied both sides by n+1?
@Salko_ but are we not proving that n! > n(n+1)/2 so using that fact in our proof is circular logic?
we multiply both sides of eq at 7:11 by n+1
@Myhair0_0 That's why we have to check whether the inequality holds for the base case (n=4 in this example) Then, the proof from the video, which @Salko_ summarized, shows that if the inequality holds for any integer >=4, then it holds for the next integer. Thus, after manually verifying that the inequality holds for n=4 and completing the short proof, we know that the inequality holds for n=4, n=4+1=5, n=5+1=6, and so on. Hope this helps
@@Salko_ I don't know why the freak my response got deleted last night, I was asking this:
Are you saying (n+1)! > (n+1)(n+2)/2 come from replacing n with n+1 in the original n! > n(n+1)/2? And does (n+1)n! > (n+1)n(n+1)/2 cone from the original but after multiplying both sides by n+1?
Hello sir,
Very interesting approach to the problem. But if we are looking for the condition of the equality happening is it not easier?
What needs to happen so n*(n+1)/2=n! ?
Since n!=0 we can divide by n so (n+1)/2=(n-1)!
n+1=2(n-1)!
n=2(n-1)!-1 If we replace n by k+1 to have a nicer number inside the factorial
k=2k!-2 so k=2(k!-1) which means that k>k!-1 or k+1>k! and we can easily verify that this proposition is only true for very small values
10:25 only if n>1, which is understood here...
n²+n isn't always bigger than n+n as it is smaller at n=(0,1)
but n is greater then 3
I know but i said that it isn't always, but it is in that case
Very rough, not rigorous idea but...
I figured that because the factorial grows much faster than the quadratic, after some point there won't be any more solutions, so there isn't any need to check every case.
I remembered the 1 + 2 + 3 = 1 x 2 x 3 meme, so n = 3
n also = 1.
That was basically my reasoning, because after n = 3 the factorial grows much faster than the quadratic so they will never ever intersect again, the only need is to check for answers within that range.
I think the proof of the factorial outlasting the triangular numbers for n >= 4 was rigorous to help cement that idea that they can never be equal and hence no solutions for that region.
Thanks!:)
Best wishes for you and your family Professor 🎉🎉😊😊❤❤
In year 2025
sir , can you record me some books to learn advanced mathematics
Begin with "Algebraic Geometry" by Hartshorne.
@Tommy_007 thank you
@@ishanpurkait9124 It's a VERY difficult book. I'll recommend that you start with basic books about calculus, elementary number theory, classical plane geometry, linear algebra, and abstract algebra.
An older series of books that are recommended for interested high school students: New Mathematical Library.
thank you ,can you help me choose between calculus by stewart ( republished by clegg and watson ) and thomas , both the early transcendental version
I immediately know of 1, and 3. I think those are the only 2 but proving that is a whole different issue. I might be able to do that but not sure exactly what I'd do maybe induction or contradiction but would take awhile to prove it.
Something bugged me,
in the induction, since we only needed n>=2
why couldn't we start at 2 ?
well, 15 years after formerly learning about induction, I finaly understand how important the initial step is (base case).
if we start at 2, initial step would be
2! = 2
2+1 = 3
2 is not greater than 3
So we can't start with 2,
And we can't start with 3 either since we have shown it is equal.
This is a marvelous Induction.
@ Prime Newtons n = 1 *OR* 3, not 1 "and" 3.
Ready for it
Hence _tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = 180°_
Can I send you a question
thank you teach us some strong induction
There are exactly 77000 ordered quadruples (a,b, c,d) such that gcd (a,b, c,d) =77 and lcm (a,b, c,d) =n, What is the smallest possible value of n?
Hello teacher. Could we look at this question? There were many solutions that i didn't understand well. I would like to see your approach
What restrictions are placed on a, b, c, d ? For example, can a, b, c, or d be a negative integer?
@yurenchu there's no restrictions I think
Let's get into the video
We need to carve a statue for this guy because he is really skilled in mathematics.
😂🎉😂🎉🎉😂
From the thumbnail: the solutions are n=1 and n=3 . For any n>3, the cumulative product is greater than and also will be increasing faster than the cumulative sum.
I'm watching the video to see if you consider n=0 a solution or not (and why).
You have n(n+1)/2 = n! iff (n + 1)/2 = (n-1)!. But one has (n - 1)! >= n - 1 > (n +1)/2 for all n > 3. So you don't have to check beyond n = 3.
I thin, the o natural number satisfying this equation are n=1 and n=3. The left side can be substituted b n*(n+1)/2 (according to gauss forula for sumof the first n natural numbers) while the rigtside is equall to n! (accordingto the definition of factorial).So we seach values for n with n*(n+1)/2=n!. Since n! is per definition n*(n-1)!, we can transform this equation to
(n+1)/2=(n-1)!. n=1 and n=3 are possiblesolutionsforthhis equation, because (1+1)/2=(1-1)! and (3+1)/2=(3-1)!. Equal numbers can not fullfill the equation, because the right side is alwas a natural number, while the left side is for even values of n neer an integer. For all odd numbers n greater than 3, (n-1)! is greater than (n+1)/2, so n=1 and n=3 are the only solutions.
For me, it was obvious,that (n-1)! is greater than (n+1)/2 for an n>3, but nice, that you gave a proof ...
n = (1;3)
Sir pls make video on fermat's last theorem ❤ lost of love from India ❤❤
*Suppose **_∃n ≥ 4 : _Σ⁽ⁿ⁾ᵢ₌₁{i} = Π⁽ⁿ⁾ᵢ₌₁{i}_*
⇒ _½n(n + 1) = n!_
⇒ _½(n + 1)!/(n - 1)! = n!_
⇒ _2n!(n - 1)! = (n + 1)!_
⇒ _2n[(n - 1)!]² = (n + 1)! = (n + 1)n(n - 1)!_
⇒ _2(n - 1)! = n + 1 = (n - 1) + 2_
⇒ _2(n - 2)! = 1 + 2/(n - 1) < 2_ since _n ≥ 4_
⇒ *_(n - 2)! < 1_** which is impossible for any factorial.*
∴ *_Σ⁽ⁿ⁾ᵢ₌₁{i} ≠ Π⁽ⁿ⁾ᵢ₌₁{i} ∀n ≥ 4_*
By inspection equality holds when *_n = 1 or 3 but not 2._*
"Satisfying" spelt wrong in the title.
For a triangular number to be equal to a factorial, we get: n(n+1)/2 = n! so n+1 = 2(n-1)! thus n is odd. Let n = 2k+1, then k+1 = (2k)! and since 2k(2k-1) ≤ k+1 means (k-1)(4k+1) ≤ 0, so k ≤ 1, we have 2k(2k-1) > k+1 for all k > 1. This concludes: (2k)! ≥ 2k(2k-1) > k+1 for k > 1, therefore k ≤ 1, leaving k = 0 and k = 1 the only options. These both lead to valid solutions in n: n = 1 and n = 3.
No, n = 1 *OR* 3.
By inspection, there is only one answer, n=3
n=1
What about n = 1?
ATTEMPT:
By inspection, N = 1, 3
1 = 1
1 + 2 + 3 = 1 * 2 * 3 = 6
N = 2 doesn't work.
1 + 2 = 3, 1 * 2 = 2
For N >= 4:
N! > N(N-1)
= N^2 - N
= (N^2)/2 + N/2 + (N^2)/2 - 3N/2
>= (N^2)/2 + N/2 + 8 - 6
> (N^2)/2 + N/2
= N(N+1)/2
Which is the sum of all natural numbers up to N.
Therefore for all natural N >= 4,
1 + 2 + 3 + ... + N < 1 * 2 * 3 * ... * N, and so the two sides cannot be equal.
The only solutions are N = 1 and N = 3.
The one slight issue is the substitution of n = 4 in the (n²-3n)/2. We would just need to show that (n²-3n)/2 is bigger than 0 for n ≥ 4. But that's easy.
n = 1 *or* 3.
Me: "Obviously its n = 1 or n = 3"
*Trying to prove that these are the ONLY values*
Me: You got me there LMAO
I thought about using induction to prove it since I saw that n = 4 didn't work and I knew for sure n > 4 didn't work either but I was a bit apprehensive knowing that it would've required some work.
No, it's n = 1 *OR* 3.
By observation: n can be 1 and 3. But not 2, 4....
No, n can be 1 *or* 3.
log( 1 + 2 + 3 ) = log(1) + log(2) + log(3)
that's actually true
Those are not natural numbers
n=1 and n=3 and in my opinion that's all possibilities
n = 1 *or* 3
@@robertveith6383 you are right, yes n=1 xor n=3
As a fact I can write you that in my language there is a word "albo" which suits the best here and it is equivalent to exclusive or which you don't use
For a, b, c, d, e ∈N
If a + b + c + d + e = abcde
Find the maximum possible value of max {a, b, c, d, e }
Sir pls explain it
You've explained it before but there are different solutions at different places and the answer got by u also don't satisfy it
I will redo it
@@PrimeNewtons thnx a lot ❤️
This is not partial texting. Spell out "You've " and "you."
@@robertveith6383 ok but it's all about conveying information 😅💗💞
my friend, please answer! ! Are you from South Africa ? I LOVE MANDELA AND BLACK PEOPLE !!!!!
Nigeria 🇳🇬
Original poster, stop yelling in all caps. A color of a person is not to be loved. That is not logical.
THE INTEGRAL.
1/(1+x⁴).
THE THIRD WAY.
WHERE IS ITTT??
Stop yelling your post in all caps.
@robertveith6383 I just wonder what the 3rd way is😞😞