Very nice approach! There is also a direct proof if one notices that k*k! = (k+1)*k! - k! = (k+1)! - k! Just like you did at the end. The initial sum becomes quite quickly (n+1)! - 1.
During watching the video, I got a similar solution also. I just add 1!+2!+...+n! to both side, then combine the right side k*k! + k! = (k+1)!, then cross out both side the same items, then come out of (n+1)! -1.
We don't need that induction, we can proove that the same way we do it in calculus. Just let c_n=a_n - b_n. Delta(c_n)=c_(n+1)-c_n=(a_(n+1)-b(n+1))-(a_n-b_n)=(a_(n+1)-a_n)-(b_(n+1)-b_n)=delta(a_n)-delta(b_n). From there, delta(c_n)=0 meaning c_n is always the same, so a_n-b_n is a constant
Finally solved 1 of these!!!! With, dare I say, a lil simpler approach. Let A(n) = 1.1! + 2.2! + ... n.n! and B(n) = 1!+2!+...+n!. A(n)+B(n) = 2.1! + 3.2! + ... (n+1).n! = 2! + 3! + ... (n+1)! = B(n+1) - 1. Then notice B(n+1) - B(n) is simply cancelling every B term expect last, or (n+1)!. Finally we combine both equations and get A(n) = B(n+1)-B(n) -1 = (n+1)! - 1.
HOMEWORK : All the sequences consisting of five letters from the set {T, U, R, N, I, P} (with repetitions allowed) are arranged in alphabetical order in a dictionary. Two sequences are called “anagrams” of each other if one can be obtained by rearranging the letters of the other. How many pairs of anagrams are there that have exactly 100 other sequences between them in the dictionary? SOURCE : 2003 HMMT - Guts Round
Not a full proof, but I believe the answer is none. The problem can be restated using base 6 numbers, to make it easier to perform math on the sequences, i.e. a sequence of five letters from a six letter set is just a five-digit base 6 number. The change in value of performing a digit swap of the xth and yth digits (denoted as d(x) and d(y) here) in a base 6 number is -d(x)*6^x + d(y)*6^x - d(y)*6^y + d(x)*6^y. With some algebra, that can be simplified to a form of (d(x)-d(y)) * (6^y-6^x). WLOG, assume x
@@codegorilla7120 SOLUTION *0* Convert each letter to a digit in base 6: I→0, N→1, P→2, R→3, T→4, U→5. Then the dictionary simply consists of all base-6 integers from 00000 to 55555 in numerical order. If one number can be obtained from another by a rearrangement of digits, then the numbers are congruent modulo 5 (this holds because a number abcde (base 6) = 6^4·a + 6^3·b + 6^2·c + 6·d + e is congruent modulo 5 to a + b + c + d + e), but if there are 100 other numbers between them, then their difference is 101, which is not divisible by 5. So there are no such pairs.
If you know about the factorial number system, then the answer just gives itself. The factorials k! play the roles of 10^k, and instead of the highest digit being 9, the highest digit for the k! place is k. If I give you 9×1+9×10+9×100+9×1000, then it's clear that's just 10000-1, because you maxed out each place, so adding 1 gives the next power of 10. Same idea here: you maxed out the digits in factorial number system up to n!, so adding 1 gives the next factorial (n+1)!, and so the final answer is (n+1)!-1.
To prove the intermediate result, you could just start from delta a_k = delta b_k and take the sum k = 1 to n-1 on both sides. It's a telescoping sum and what's left is a_n - a_1 = b_n - b_1, thus a_n = b_n + some constant. No need for induction.
@@schweinmachtbree1013 well to be honest the whole problem can be solved by a 10-second telescoping sum, as it has been suggested in the comments. He perhaps wanted to show the "experimentation" required when tackling these calculations.
This is one of my favorite videos from him. It solves a hard problem, and introduces the ideas of discreate derivative in an easy to follow manner. Anytime anything mathematical can be presented in an easy to follow manner is Instructional Genius :)
Another way that does not requrie any calculus knowledge, is to split k*k! to (k+1)!-k!. When we staet writing out the series, it goes like this: Sn=(2! - 1!) + (3! - 2!) + (4! - 3!) + (5! - 4!) +(6! - 5!) + (7! - 6!) + ... + (n! - (n-1)!) + ((n+1)! - n!). From here it is trivial to see that every term exept (n+1)! and -1 cansels, so the answer is : (n+1)!-1
You can use the gamma function to get a telescoping sum, we know that: Γ(k+2)=(k+1)Γ(k+1) =>Γ(k+2)-Γ(k+1)=kΓ(k+1) so Ʃk.k!=ƩkΓ(k+1)=ƩΓ(k+2)-Γ(k+1)=Γ(n+2)-Γ(2)=(n+1)!-1
This video's philosophy: Why do something in 30 seconds if you can find a way to do it in ten minutes? It's not unreasonable to notice that n*n! = (n+1)! - n!.
Nice solution! Just like Michael mentioned: if you are at a contest just try evaluating the expression for small values (that is guessing the answer) Of course you should expect tha answer to contain factorial and n+1... Then just prove your guess by induction! Alternative: you can also try doing some telescopic sums ...
Took me a while to realize that n*n! = (n+1)! - n! and if we sum it up for the whole series, (n+1)! - 1 is exactly what remains. This way it feels much more straightforward than guessing from first few values, too, and I guess doesn't require induction to prove, though it's always an option.
it is certainly much quicker to do it your way, but if you look at the few values and guess that the sum is (n+1)! - 1, you might spot that this is (n+1)! - 1! which looks like the answer to a telescoping sum, which might cause you to notice the alternate expression n*n! = (n+1)! - n!, and then simply perform the telescoping sum.
A helpful set of data to guess by has headings n, n!, aₙ = n.n! and Sₙ. With the n! column, it is glaringly obvious. With just the aₙ and Sₙ, it's much less clear. That's how I found my guess, anyway: your mileage may vary, as they say.
Found the simplest solution: Let S = 1.1! + 2.2! + ...+ n.n! And Sn = 1! + 2! + 3! +... +n! Adding both, S+ Sn = 2.1! + 3.2! +... (n+1). n! =2! + 3! +... +(n+1)! S + Sn = Sn + (n+1)!-1! =>S = (n+1)!-1
@@sharathpr42 Fancy characters are a pain to type but sometimes there are possibilities: ₙ is U+2099. You might have a way to type that with the keypad or something. On Android, there are apps such as Character Pad, from which I pasted it. So Sₙ is possible, but a right PITA.
Another solution: Let a(n)=this sum, let f(n)=sum of n first factorials. a(n)+f(n) = sum from k=1 to n of k*k! + sum from k=1 to n of k! = sum from k=1 to n of (k+1)*k! = sum from k=1 to n of (k+1)! = f(n+1)-1. Therefore a(n)=f(n+1)-f(n)-1=(n+1)!-1.
Hi Michael It can be shown easily without induction by using gamma function together with its first well-known identity which is gamma(i+z)=z times gamma(z)and reindexing indices , I’ve solved it in three lines. Have a nice day!
A combinatorial proof: We want to count the permutations of n+1 elements, that do not fix at least one element. We do this by partitioning these permutations into n+1 sets and counting them separately: Let the k-th set contain all permutations that dont fix the k-th element, but fix all elements from k+1 to n+1. These sets don't overlap and contain all permutations, since for every non-constant permutation, there always exists a unique k in {1,2,...,n+1} such that it isn't fixed and everything bigger is. The k=1 set is empty, since there is no permutation fixing everything except for one element. The k=2 set has only one element, namely the unique permutation exchanging 1 and 2 and leaving everything else fixed, etc. In general, the k-th set contains (k-1)*(k-1)! permutations, since there are k-1 choices for the image of the k-th element, only one choice for alla elements bigger than k and then (k-1)! choices for the first k-1 elements. This means that: Σ(k-1)(k-1)!=(n+1)!-1, where the sum is for k=1 to n+1, which is the desired equation.
I did it using k!=integral t= 0 to inf (exp(-t) t^k dt. It's fairly easy to take the sum inside the integral and do a few standard tricks with integration by parts. Still harder than guess+simple induction. Doh!
Nice method to answer without induction or using series: If you take x=1.1!+2.2!+...n.n! then adding (1!+2!+..+n!) leads to: x + (1!+2!+..+n!) = (2!+3!+..+(n+1)!) Almost everything cancels out so x= (n+1)!-1
It’s cool you can make a number system with factorials. c[1]*1! +c[2]*2! +...c[n]*n! is one-to-one and onto with the numbers 1..(n+1)!-1 as long as the coefficients are 0
This is an interesting method. Though its a lot quicker to observe that the sum Sn we want to derive and Un the sum of the n first factorials verify : Sn + Un = U(n+1) - 1. The resulting telescoping sum gives us the solution directly.
I think we still have to “spot the solution” to proceed as Michael did. Had we not been able to spot a closed form for b_n, we would be back to square one. All in all I think there is no difference in mathematical content between this solution and the conventional one.
@@failsmichael2542 well you can evaluate the sum directly without knowing/guessing the answer first by spotting that it is a telescoping sum: sum(k k!) = sum((k+1 - 1)k!) = sum((k+1)k! - k!) = sum((k+1)! - k!), which telescopes to (n+1)! - 1 as required.
My working on this puzzle was very similar, but I approached the "look at it and try to intuit a guess for the formula, before going on to prove it" stage from the opposite direction: To start with, n*n! is kinda a weird object, it's a weird expression to be working with. It would be much nicer if it was (n+1)*n! instead, since that equals (n+1)!. The difference between n*n! and (n+1)*n! is just n!, so it would be nice if all the other terms of our sum added up to n!, so that when we add it to n*n! we get (n+1)!. That sure smells like something that could be turned into an inductive step. So, our first guess is that a_n = (n+1)!, it fits the inductive case, but unfortunately it doesn't fit the base case: a_n = 1, not 2. But that's not a major issue, as our intuctive step is only really looking at the new term that's been added onto the sum, so it should be resilient to simply adding or subtracting a constant to all the terms (my reasoning here is intuitive and handwavy, but a rigorous version of it is essentially the discrete-derivative property shown in the video). So our new guess is a_n = (n+1)! - 1... that fits our base case, and should still fit our inductive case. And then the proof continues in the normal fashion from there.
In my opinion the easier method is: Write 2.2! as 3.2! - 1.2! then 3.2! is obviously 3! combine that with 3.3! and you get 4.3! which is 4!. Combine that with 4.4! which is 5! and so on and on until you get to (n+1).n! which is n+1!. The remaining two terms are 1! and -2! which give -1. So all adds up to (n+1)! -1. Or nicer written as a telescopic sum: sum_k=1^n k.k!=sum_k=1^n [(k+1)k!-k!]=sum_k=1^n (k+1)! - sum_k=1^n k! = sum_k=2^n+1 k! - sum_k=1^n k! = (n+1)! - 1!
nice, so it is only : (n+1)! = (n+1) n! = n*n! + n! = n*n! + (n-1)!*(n-1)+ (n-1)! =... =n*n! + (n-1)!*(n-1)+... + 2*2! +2! = n*n! + (n-1)!*(n-1)+... + 2*2! +1*1! + 1 or shorter: (n+1)!-n! =(n+1)n! - n! = (n+1-1) n! = n n! then replace all n*n! as (n+1)! - n! you will get telescoping row
So happy and surprised to see this problem on Michael's channel. I've recently found this problem in algebra task book for 8 or 9 graders (15 - 16 y. o. ) from advanced math classes. That book is written by M. Galitsky. (Сборник задач по алгебре. Учебное пособие для 8 - 9 классов с углубленным изучением математики)
I followed along with every single step. But I have no idea why we just did this entire thing haha. This is what I get for being an engineer and not a pure mathematician ...
A job well done, but needlessly lengthened. A simple way would be to write all k.k! = (k+1)! - k!, then write the required sum as a difference of two sums, cancel common terms and we'll be left with (n+1)! - 1 directly.
Yep you can easilly derive the formula for 1!+2!+...n! using facts from the video and then just take the diferences if you want start to adding factorials from n! end to k!
Given aₙ=ⁿ∑ᵢ₌₁i*i! i*i! almost looks like (i+1)*i! which is (i+1)! So i*i!+i!-i!=(i+1)*i!-i!=(i+1)!-i! Then aₙ=ⁿ∑ᵢ₌₁((i+1)!-i!)=ⁿ∑ᵢ₌₁((i+1)!)-ⁿ∑ᵢ₌₁(i!) If we reindex tgen aₙ=ⁿ⁺¹∑ᵢ₌₂(i!)-ⁿ∑ᵢ₌₁(i!)=(ⁿ∑ᵢ₌₂(i!)+(n+1)!)-(ⁿ∑ᵢ₌₂(i!)+1!) The sums cancel so aₙ=(n+1)!-1
You can use the fact that n! +n.n! =(n+1) ! S=1.1! +2.2! +3.3! +..... +n.n! S+1=1! +1.1! +......... +n.n! Using above fact 1! +1.1! Becomes 2! 2! +2.2! Becomes 3! And soon It becomes (n+1) ! S=(n+1)! -1
Without using calculus we can solve this by a simple trick.. We will rewrite: k = (k+1)-1 for all k=1,2,3,...,n Now, 1.1!+2.2!+...+n.n! = (2-1).1!+(3-1).2!+...+((n+1)-1).n! = (2!+3!+...+(n+1)!)-(1!+2!+...+n!) = (n+1)!-1!
I have got a very very nice method for this problem . And want to share it with you I can't type that in comment box so can you just give your mail id?
Actually one doesn't have to show the methodology in JEE exams. It can be easily done with the standard method. Michael sir finds the solution using a unique method.
@@adithyan9263 Yeah we are solving problems because we love math over here unlike a certain exam which wants to make us machines by solving problems without seeing the beauty in the subject itself.
Another well produced and well presented video. Michael Penn keeps raising the bar for maths on TH-cam.
Tbh his videos’ quality dropped significantly
@@nurtasshyntas7745 L
Very nice approach!
There is also a direct proof if one notices that k*k! = (k+1)*k! - k! = (k+1)! - k! Just like you did at the end. The initial sum becomes quite quickly (n+1)! - 1.
Nice! Came here to comment this, but you beat me to it :p
During watching the video, I got a similar solution also. I just add 1!+2!+...+n! to both side, then combine the right side k*k! + k! = (k+1)!, then cross out both side the same items, then come out of (n+1)! -1.
Yeah buddy I thought the same
Yes, it 'telescopes'.
Yep
We don't need that induction, we can proove that the same way we do it in calculus. Just let c_n=a_n - b_n. Delta(c_n)=c_(n+1)-c_n=(a_(n+1)-b(n+1))-(a_n-b_n)=(a_(n+1)-a_n)-(b_(n+1)-b_n)=delta(a_n)-delta(b_n). From there, delta(c_n)=0 meaning c_n is always the same, so a_n-b_n is a constant
Finally solved 1 of these!!!! With, dare I say, a lil simpler approach.
Let A(n) = 1.1! + 2.2! + ... n.n! and
B(n) = 1!+2!+...+n!.
A(n)+B(n) = 2.1! + 3.2! + ... (n+1).n! = 2! + 3! + ... (n+1)! = B(n+1) - 1.
Then notice B(n+1) - B(n) is simply cancelling every B term expect last, or (n+1)!.
Finally we combine both equations and get A(n) = B(n+1)-B(n) -1 = (n+1)! - 1.
This sequence , expressed as the sum above, is A033312 in the Online Encyclopedia of Integer Sequences.
9:07 Michael’s homework
13:55 Good Place To Stop
how are you always this early? XD
Why do you do this everytime
HOMEWORK : All the sequences consisting of five letters from the set {T, U, R, N, I, P} (with repetitions allowed) are arranged in alphabetical order in a dictionary. Two sequences are called “anagrams” of each other if one can be obtained by rearranging the letters of the other. How many pairs of anagrams are there that have exactly 100 other sequences between them in the dictionary?
SOURCE : 2003 HMMT - Guts Round
That's a pretty hard homework . Atleast for me
Not a full proof, but I believe the answer is none. The problem can be restated using base 6 numbers, to make it easier to perform math on the sequences, i.e. a sequence of five letters from a six letter set is just a five-digit base 6 number. The change in value of performing a digit swap of the xth and yth digits (denoted as d(x) and d(y) here) in a base 6 number is -d(x)*6^x + d(y)*6^x - d(y)*6^y + d(x)*6^y. With some algebra, that can be simplified to a form of (d(x)-d(y)) * (6^y-6^x). WLOG, assume x
@@codegorilla7120 SOLUTION
*0*
Convert each letter to a digit in base 6: I→0, N→1, P→2, R→3, T→4, U→5. Then the dictionary simply consists of all base-6 integers from 00000 to 55555 in numerical order. If one number can be obtained from another by a rearrangement of digits, then the numbers are congruent modulo 5 (this holds because a number abcde (base 6) = 6^4·a + 6^3·b + 6^2·c + 6·d + e is congruent modulo 5 to a + b + c + d + e), but if there are 100 other numbers between them, then their difference is 101, which is not divisible by 5. So there are no such pairs.
Wow what a cool solution
I tried but couldnt do it
Never thought of using base 6🤣
Thnx for the novel approach
This is also an interesting problem if you're asked how many pairs are arranged such that the second is the 100th item counting from the first.
If you know about the factorial number system, then the answer just gives itself.
The factorials k! play the roles of 10^k, and instead of the highest digit being 9, the highest digit for the k! place is k.
If I give you 9×1+9×10+9×100+9×1000, then it's clear that's just 10000-1, because you maxed out each place, so adding 1 gives the next power of 10. Same idea here: you maxed out the digits in factorial number system up to n!, so adding 1 gives the next factorial (n+1)!, and so the final answer is (n+1)!-1.
To prove the intermediate result, you could just start from delta a_k = delta b_k and take the sum k = 1 to n-1 on both sides. It's a telescoping sum and what's left is a_n - a_1 = b_n - b_1, thus a_n = b_n + some constant.
No need for induction.
brilliant!
@@schweinmachtbree1013 well to be honest the whole problem can be solved by a 10-second telescoping sum, as it has been suggested in the comments. He perhaps wanted to show the "experimentation" required when tackling these calculations.
Well produced and expertly narrated. You’re amazing Michael!
This is one of my favorite videos from him. It solves a hard problem, and introduces the ideas of discreate derivative in an easy to follow manner. Anytime anything mathematical can be presented in an easy to follow manner is Instructional Genius :)
Another way that does not requrie any calculus knowledge, is to split k*k! to (k+1)!-k!. When we staet writing out the series, it goes like this:
Sn=(2! - 1!) + (3! - 2!) + (4! - 3!) + (5! - 4!) +(6! - 5!) + (7! - 6!) + ... + (n! - (n-1)!) + ((n+1)! - n!). From here it is trivial to see that every term exept (n+1)! and -1 cansels, so the answer is : (n+1)!-1
You can use the gamma function to get a telescoping sum, we know that:
Γ(k+2)=(k+1)Γ(k+1)
=>Γ(k+2)-Γ(k+1)=kΓ(k+1)
so
Ʃk.k!=ƩkΓ(k+1)=ƩΓ(k+2)-Γ(k+1)=Γ(n+2)-Γ(2)=(n+1)!-1
U could have also just used the factorial instead of the gamma function
This video's philosophy: Why do something in 30 seconds if you can find a way to do it in ten minutes? It's not unreasonable to notice that n*n! = (n+1)! - n!.
If I have the time, I will spend an hour automating something that can be done in 5 minutes. It's good to practice.
Telescope sum, nice
I get that the intention is to approach the problem from the standpoint of calculus but frankly this is an overkill 😂
More than an overkill, it shows that there's some powerful theory on the background to expolit.
Nice solution!
Just like Michael mentioned: if you are at a contest just try evaluating the expression for small values (that is guessing the answer)
Of course you should expect tha answer to contain factorial and n+1...
Then just prove your guess by induction!
Alternative: you can also try doing some telescopic sums ...
Took me a while to realize that n*n! = (n+1)! - n! and if we sum it up for the whole series, (n+1)! - 1 is exactly what remains. This way it feels much more straightforward than guessing from first few values, too, and I guess doesn't require induction to prove, though it's always an option.
it is certainly much quicker to do it your way, but if you look at the few values and guess that the sum is (n+1)! - 1, you might spot that this is (n+1)! - 1! which looks like the answer to a telescoping sum, which might cause you to notice the alternate expression n*n! = (n+1)! - n!, and then simply perform the telescoping sum.
A helpful set of data to guess by has headings n, n!, aₙ = n.n! and Sₙ. With the n! column, it is glaringly obvious. With just the aₙ and Sₙ, it's much less clear. That's how I found my guess, anyway: your mileage may vary, as they say.
If you've already guessed (n+1)!-1, you can prove it with induction in a heartbeat, no discrete calculus is needed.
i really like discrete calculus, it makes some sequence/series problems really smooth
Found the simplest solution:
Let S = 1.1! + 2.2! + ...+ n.n!
And Sn = 1! + 2! + 3! +... +n!
Adding both,
S+ Sn = 2.1! + 3.2! +... (n+1). n!
=2! + 3! +... +(n+1)!
S + Sn = Sn + (n+1)!-1!
=>S = (n+1)!-1
very nice, but I'd write "T" instead of "Sn" to avoid the possible confusion that "Sn" means "S*n"
@@schweinmachtbree1013 thanks.. couldn't type it as a subscript.. so I can see how it can look confusing.
@@sharathpr42 Fancy characters are a pain to type but sometimes there are possibilities: ₙ is U+2099. You might have a way to type that with the keypad or something. On Android, there are apps such as Character Pad, from which I pasted it. So Sₙ is possible, but a right PITA.
Another solution:
Let a(n)=this sum, let f(n)=sum of n first factorials.
a(n)+f(n) = sum from k=1 to n of k*k! + sum from k=1 to n of k! = sum from k=1 to n of (k+1)*k! = sum from k=1 to n of (k+1)! = f(n+1)-1.
Therefore a(n)=f(n+1)-f(n)-1=(n+1)!-1.
Hi Michael
It can be shown easily without induction by using gamma function together with its first well-known identity which is gamma(i+z)=z times gamma(z)and reindexing indices , I’ve solved it in three lines.
Have a nice day!
A combinatorial proof:
We want to count the permutations of n+1 elements, that do not fix at least one element. We do this by partitioning these permutations into n+1 sets and counting them separately: Let the k-th set contain all permutations that dont fix the k-th element, but fix all elements from k+1 to n+1. These sets don't overlap and contain all permutations, since for every non-constant permutation, there always exists a unique k in {1,2,...,n+1} such that it isn't fixed and everything bigger is. The k=1 set is empty, since there is no permutation fixing everything except for one element. The k=2 set has only one element, namely the unique permutation exchanging 1 and 2 and leaving everything else fixed, etc. In general, the k-th set contains (k-1)*(k-1)! permutations, since there are k-1 choices for the image of the k-th element, only one choice for alla elements bigger than k and then (k-1)! choices for the first k-1 elements. This means that:
Σ(k-1)(k-1)!=(n+1)!-1, where the sum is for k=1 to n+1, which is the desired equation.
(Which also explains intuitively where is the term 0*0! and why this -1 on the right side is not in the form k*k!, when we bring it to the left side)
This looks like Proof by Induction but with extra steps.
I did it using k!=integral t= 0 to inf (exp(-t) t^k dt. It's fairly easy to take the sum inside the integral and do a few standard tricks with integration by parts. Still harder than guess+simple induction. Doh!
This was the first thing that I tried, too.
Thank you for the idea with integration by parts, I had no idea how to go on :)
Nice method to answer without induction or using series:
If you take x=1.1!+2.2!+...n.n! then adding (1!+2!+..+n!) leads to:
x + (1!+2!+..+n!) = (2!+3!+..+(n+1)!)
Almost everything cancels out so
x= (n+1)!-1
It’s cool you can make a number system with factorials. c[1]*1! +c[2]*2! +...c[n]*n! is one-to-one and onto with the numbers 1..(n+1)!-1 as long as the coefficients are 0
This is an interesting method. Though its a lot quicker to observe that the sum Sn we want to derive and Un the sum of the n first factorials verify : Sn + Un = U(n+1) - 1. The resulting telescoping sum gives us the solution directly.
Once you spot the solution, you can prove it directly pretty easily. Though you don't get the fun diversion through discrete differentiation that way
I think we still have to “spot the solution” to proceed as Michael did. Had we not been able to spot a closed form for b_n, we would be back to square one. All in all I think there is no difference in mathematical content between this solution and the conventional one.
@@failsmichael2542 well you can evaluate the sum directly without knowing/guessing the answer first by spotting that it is a telescoping sum: sum(k k!) = sum((k+1 - 1)k!) = sum((k+1)k! - k!) = sum((k+1)! - k!), which telescopes to (n+1)! - 1 as required.
@@schweinmachtbree1013 Indeed. I wanted to compare Michael's solution to the usual induction proof (i.e guesswork :)) Sorry for not being clear.
My working on this puzzle was very similar, but I approached the "look at it and try to intuit a guess for the formula, before going on to prove it" stage from the opposite direction:
To start with, n*n! is kinda a weird object, it's a weird expression to be working with. It would be much nicer if it was (n+1)*n! instead, since that equals (n+1)!. The difference between n*n! and (n+1)*n! is just n!, so it would be nice if all the other terms of our sum added up to n!, so that when we add it to n*n! we get (n+1)!. That sure smells like something that could be turned into an inductive step.
So, our first guess is that a_n = (n+1)!, it fits the inductive case, but unfortunately it doesn't fit the base case: a_n = 1, not 2. But that's not a major issue, as our intuctive step is only really looking at the new term that's been added onto the sum, so it should be resilient to simply adding or subtracting a constant to all the terms (my reasoning here is intuitive and handwavy, but a rigorous version of it is essentially the discrete-derivative property shown in the video). So our new guess is a_n = (n+1)! - 1... that fits our base case, and should still fit our inductive case.
And then the proof continues in the normal fashion from there.
In my opinion the easier method is: Write 2.2! as 3.2! - 1.2! then 3.2! is obviously 3! combine that with 3.3! and you get 4.3! which is 4!. Combine that with 4.4! which is 5! and so on and on until you get to (n+1).n! which is n+1!. The remaining two terms are 1! and -2! which give -1. So all adds up to (n+1)! -1.
Or nicer written as a telescopic sum: sum_k=1^n k.k!=sum_k=1^n [(k+1)k!-k!]=sum_k=1^n (k+1)! - sum_k=1^n k! = sum_k=2^n+1 k! - sum_k=1^n k! = (n+1)! - 1!
can we use the gamma function here?
YESSSS FINALLY MORE DISCRETE CALC!!!! Please upload more discrete calc, or post resources, i’m desperate!!
I am happy Michael did well in this one. Well explained, not too fast, perfect.
Additionally, if you were to add 0! so that a(0)=0!=1 and a(n+1)=a(n)+(n+1)(n+1)! then a(n)=n!
no 0.0!=0 so a(0)=0
Hello sir, can we use gamma function to do this??
maybe, but if you can it definitely isn't easier.
nice, so it is only : (n+1)! = (n+1) n! = n*n! + n! = n*n! + (n-1)!*(n-1)+ (n-1)! =... =n*n! + (n-1)!*(n-1)+... + 2*2! +2! = n*n! + (n-1)!*(n-1)+... + 2*2! +1*1! + 1
or shorter:
(n+1)!-n! =(n+1)n! - n! = (n+1-1) n! = n n!
then replace all n*n! as (n+1)! - n! you will get telescoping row
So happy and surprised to see this problem on Michael's channel. I've recently found this problem in algebra task book for 8 or 9 graders (15 - 16 y. o. ) from advanced math classes. That book is written by M. Galitsky. (Сборник задач по алгебре. Учебное пособие для 8 - 9 классов с углубленным изучением математики)
Thanks. Could you solve this with help of generation functions and combinatorial argument?
I followed along with every single step. But I have no idea why we just did this entire thing haha. This is what I get for being an engineer and not a pure mathematician ...
Good Place To Start at 9:41
A job well done, but needlessly lengthened. A simple way would be to write all k.k! = (k+1)! - k!, then write the required sum as a difference of two sums, cancel common terms and we'll be left with (n+1)! - 1 directly.
S=(1/2)*2!+(2/3)*3!+...+(n/(n+1))*(n+1)!= Suma(k!)*((k-1)/k)= Suma(k!)-Suma((k-1)!) where suma go from k=2 till n+1
Does exist formula for n!+(n+1)!+(n+2)!+(n+3)!...?
Yep you can easilly derive the formula for 1!+2!+...n! using facts from the video and then just take the diferences if you want start to adding factorials from n! end to k!
Flammable math already done it with an ez method
سلام عليكم انت عربي؟
@@user-engahmed نعم
@@youssefsahraoui4802 ياهلا ومرحبآ.. كيف بتفهم على الشرح يوجد ترجمة للفيدوات؟
Just write n.n!=(n+1-1).n!=(n+1)!-n!. Then telescope... BTW, I have learned new things through you..
So... (n+1)! = n * n! + (n-1) * (n-1)! + . . . + 2 * 2! + 1 * 1! + 0!, right?
yeah
Given aₙ=ⁿ∑ᵢ₌₁i*i!
i*i! almost looks like (i+1)*i! which is (i+1)!
So i*i!+i!-i!=(i+1)*i!-i!=(i+1)!-i!
Then aₙ=ⁿ∑ᵢ₌₁((i+1)!-i!)=ⁿ∑ᵢ₌₁((i+1)!)-ⁿ∑ᵢ₌₁(i!)
If we reindex tgen
aₙ=ⁿ⁺¹∑ᵢ₌₂(i!)-ⁿ∑ᵢ₌₁(i!)=(ⁿ∑ᵢ₌₂(i!)+(n+1)!)-(ⁿ∑ᵢ₌₂(i!)+1!)
The sums cancel so aₙ=(n+1)!-1
You can use the fact that
n! +n.n! =(n+1) !
S=1.1! +2.2! +3.3! +..... +n.n!
S+1=1! +1.1! +......... +n.n!
Using above fact
1! +1.1! Becomes 2!
2! +2.2! Becomes 3! And soon
It becomes (n+1) !
S=(n+1)! -1
I am surprised
Without using calculus we can solve this by a simple trick..
We will rewrite:
k = (k+1)-1 for all k=1,2,3,...,n
Now,
1.1!+2.2!+...+n.n!
= (2-1).1!+(3-1).2!+...+((n+1)-1).n!
= (2!+3!+...+(n+1)!)-(1!+2!+...+n!)
= (n+1)!-1!
Nice but Overkill
wow an easy one. makes me feel slightly less dumb after all your other videos ive seen.
Today, we won't solve the problem by induction. Instead we will use a foreign, convoluted lemma and prove it... by induction. 😂
This is an IITJee question from India.
X(X!) = (X+1)!-X!
So if we do this to all, everything will get cancelled out and we get (n+1)!-1
please make a video about how we dont know whether pi+e is rational or not, and how pathetic that is
Is it just me or he looks a little like Neil Patrick Harris
k*k!=(k+1-1)*k!=(k+1)*k!-k!=(k+1)!-k!
And after that easy solution. 😉
Using linearity you can find the result pretty quickly
what's that?
How?
Linearity of what?
@@NoNameAtAll2
k*k!
=(k+1-1)k!
=(k+1)k! -k!
=(k+1)! -k!
So summation from k=1ton
=2! -1! +3! -2! +...+(n+1)! -n!
Which telescopes to
*(n+1)! -1*
@@nicolodangelo45you can eassilly prove the linearity of the dretzen function and then take the sums
I hope to put a translation into Arabic, please, please🙏🥺
Mathematics is an universal language
I have got a very very nice method for this problem . And want to share it with you I can't type that in comment box so can you just give your mail id?
Pls. Real men use Gosper's algorithm.
Or I rather guess real computers to that.
Simple question for jee aspirants from india
Actually one doesn't have to show the methodology in JEE exams. It can be easily done with the standard method.
Michael sir finds the solution using a unique method.
most jee aspirants doesn't know about discrete differentiation.
and the objective wasn't to show 1.1!+2.2!+3.3!+...n.n!=(n+1)!-1
so shut up.
@@adithyan9263 yes true.
Can you please shut up about jee for a minute.
@@adithyan9263 Yeah we are solving problems because we love math over here unlike a certain exam which wants to make us machines by solving problems without seeing the beauty in the subject itself.