Make it look like a simple calculus problem.

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[benjaminbrat]

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.

[henrymarkson]

Another well produced and well presented video. Michael Penn keeps raising the bar for maths on YouTube.

[thisisnotmyrealname]

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. From there, delta(c_n)=0 meaning c_n is always the same, so a_n-b_n is a constant

[goodplacetostop]

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

[MrPejotah]

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.

[alnitaka]

This sequence, expressed as the sum above, is A033312 in the Online Encyclopedia of Integer Sequences.

[goodplacetostop]

9:07 Michael’s homework
13:55 Good Place To Stop

[mokkapatisiddharth]

I get that the intention is to approach the problem from the standpoint of calculus but frankly this is an overkill 😂

[athysw.e.]

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.

[Jon]

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 :)

[michaelz]

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!.

[potawatomi]

Well produced and expertly narrated. You’re amazing Michael!

[littlefermat]

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 ...

[demenion]

i really like discrete calculus, it makes some sequence/series problems really smooth

[kasuha]

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.

[f-th]

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.

[wesleydeng]

If you've already guessed (n+1)!-1, you can prove it with induction in a heartbeat, no discrete calculus is needed.

[aleksandervadla]

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

[عمرانآلعمران-وخ]

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!

[michaelslack]

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