Hi sir,I am a JEE aspirant and your videos help a lot and I love your teaching style you make the hardest questions look so easy. I just had a request that please make more videos on permutation and combinations🙏🙏. I like your hat by the way
again finding last digits of sth.? ok lets go! 0:41 no that is no option it is to big and cant we simplify that? 0:48 good question next question! 1:01 ok plese give me that as formular my english seems to be to bad i don´t understand that what is the quabe of a number? 1:24 a ok so the sum of qubes from 0 to n is the same as the square of the sume from 0 to n ok that is a grate formulat! ok but from here it is not so complicate the sum of 0 to n is (n+n²)/2 and the squar an the squarroot cross each other out so whats left is the last digit of (2024+2024²)/2 = 2.049.300 and the last tree digits are 300 finish! am i right? 2:31 that symbol is quite nice please use it more often instead of 1 +... + n or sth. like this 3:33 thats what i thought 3:59 yes it is a realy useful formular! 4:10 less you but you dont use here just n+n² / 2? 5:01 oh ok because you want to explain it ok 7:04 if i was right it should be sth. ending with 300 7:10 so it works i knew sth. yes! and it was not so complicate! 7:23 it should be 2.049.300 but ok 7:28 that is the time were that bekomme my first thought after waking up :D 7:34 ok the last sentencess do not make as much sense for me as the rest but ok! bye LG K.Furry
Indeed. These types of problems are artificially created so they can be solved. Most real life problems need brute force computational methods to solve them.
A bit disappointing that all the trivial things (like the sum of consecutive integers) are explained in depth, but the only interesting thing (sum of cubes is square of sum) is only stated and restated, but not explained.
For an in depth explanation of the sum of powers, I recommend the Mathologer video on Power Sums. I find that generally, I cannot keep up with Mathologer videos but I find that the chapter on algebra at around 15 minutes is not too hard. It explains how the sum of i^n can be defined in terms of the sums of i^0, i^1, i^(n-1) using the binomial expansion of the expression (x-1)^n. The fact that the square of the sum equals the sum of the cubes sort of falls of the equations.
Hi sir,I am a JEE aspirant and your videos help a lot and I love your teaching style you make the hardest questions look so easy. I just had a request that please make more videos on permutation and combinations🙏🙏. I like your hat by the way
Mee to bro ..
Do happy to see an Indian(especially jee aspirant)😊😊
@@sajuvasu me too bro
Best of luck for jee ❤
Me too
@@Zerotoinfinityroad thanks bro same to you
@@sriraghavapabbaraja8641 all the best in jee bro
Funny that only the last three digits are asked when it is so easy to evaluate the whole number (2,049,300) if you know how to solve this problem.
Sir, you explain it very simply, I love your account so much, loves..😊❤❤
Thanks
Thank you. Deeply appreciated.
Wow that's amazing 🤩
again finding last digits of sth.? ok lets go!
0:41 no that is no option it is to big and cant we simplify that?
0:48 good question next question!
1:01 ok plese give me that as formular my english seems to be to bad i don´t understand that what is the quabe of a number?
1:24 a ok so the sum of qubes from 0 to n is the same as the square of the sume from 0 to n ok that is a grate formulat!
ok but from here it is not so complicate the sum of 0 to n is (n+n²)/2 and the squar an the squarroot cross each other out so whats left is
the last digit of (2024+2024²)/2 = 2.049.300 and the last tree digits are 300 finish! am i right?
2:31 that symbol is quite nice please use it more often instead of 1 +... + n or sth. like this
3:33 thats what i thought
3:59 yes it is a realy useful formular!
4:10 less you but you dont use here just n+n² / 2?
5:01 oh ok because you want to explain it ok
7:04 if i was right it should be sth. ending with 300
7:10 so it works i knew sth. yes! and it was not so complicate!
7:23 it should be 2.049.300 but ok
7:28 that is the time were that bekomme my first thought after waking up :D
7:34 ok the last sentencess do not make as much sense for me as the rest but ok!
bye
LG K.Furry
Another great narration
@@PrimeNewtons you are realy welcome!
but i have one question do you recived my mail?
yours sincerly
KPunktFurry
Nice
Sum of first n natural numbers - n*(n+1)/2
really great
Simply square rooting the sum of cubes up to n formula gives the sum of square so (1 + 2024)/(2024 / 2) = 2049300 so 300.
Well done
o,mj i didnt know that thanks
The last 3 digits are 300, and the full answer is 2049300. I was able to calculate it without using calculator or paper.
Another easy question dressed up to look hard.
The question itself is easy enough, but it's interesting to think that just adding 1 to the inside makes the question impossible.
Indeed. These types of problems are artificially created so they can be solved. Most real life problems need brute force computational methods to solve them.
No, it not "interesting," because then it would be a faulty problem which the instructor would not give in the first place.
A bit disappointing that all the trivial things (like the sum of consecutive integers) are explained in depth, but the only interesting thing (sum of cubes is square of sum) is only stated and restated, but not explained.
There is other information on the Internet.
He already made a video about that.
@@Rednodge_9 Ah, indeed. And it's even linked in the description. 😣
For an in depth explanation of the sum of powers, I recommend the Mathologer video on Power Sums. I find that generally, I cannot keep up with Mathologer videos but I find that the chapter on algebra at around 15 minutes is not too hard. It explains how the sum of i^n can be defined in terms of the sums of i^0, i^1, i^(n-1) using the binomial expansion of the expression (x-1)^n. The fact that the square of the sum equals the sum of the cubes sort of falls of the equations.
Plz make a proof video on "e" and "pie" are irrational😊
The product of pi and e may be written as pi × e, pi(e), pi*e, for example, but not pie.