Two nice series from India and Morocco
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- Опубліковано 25 чер 2024
- Can you solve these fun problems? Problem 1, thanks to Kanad for the suggestion!
0:00 problems
1:31 solution 1
4:36 solution 2
Problem 1 BYJU's
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Problem 2 Tahiri Math Science Online Teaching
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• Moroccan Mathematical ...
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I solved the second one slightly differently, although it still served as a telescoping solution. I expanded the numerator in each term to get (x^2 - 1)/(x^2), which factored into (x-1)(x+1)/(x^2). The vast majority of the terms in the numerator and denominator cancelled out; for example, the (3^2) in the denominator cancelled out with the (2+1) and (4-1) in the numerator. After all the cancellations, you're left with (2-1)(2023+1)/(2*2023), which reduces to the same answer you got.
The first one is indeed very famous in India, I've seen it thrice in the course of a month in different books. For context, I'm a highschooler.
Was quite fun to work out.
IIT asked this years ago, any aspirant can just look at the question and ans directly if he knows the trick.
I agree
These type of sums always have substitution of x-1,x-2 or 2x etc
Basically simple expressions are to be substituted
Yeah man, didn't even need to lift a pen for the first one. They are both standard problems.
True lol
Yeah man I didn't even have to be born to solve this cuz I already did it like lifetimes ago, right?
Yeah it's everywhere around here lol
Love from India ❤🙏
for question 2 i did it in this way-
Let S be that product then,
ln(s)= summation of ln[1-1/(r^2)] where r varies from 2 to 2023.
Now, [1-1/(r^2)] = [(r^2-1)/r^2] = [(r+1)(r-1)]/r^2
implies, ln[1-1/(r^2)]= ln[(r+1)(r-1)/r^2] = [ln(r+1) - ln r] + [ln(r-1) - lnr]
since lnr^2= 2ln r.
above sum is telescoping series hence you will get s=1012/2023
The second one is quite classic in morocco
I solved the second problem easily, but the first was a challenge. Thank you though for the difference of square explanation!
I was the opposite. I got the first one just fine but couldn't get the second one. I slapped my forehead when he showed the simplification.
@@semvhu1494 same here, first was easy, for second one i thought of a2-b2 as well but had no clue how to proceed further 😅
We love preshtalwalker that’s all
Honestly, very fun series questions
for question 2 i used geometric series for odd numbers and whatnot for alternating numbers(3/4,24/25,48/49 and so on)which led to nothing
Thanks for the solution Presh!!
Thank you for your shareing Master
The first problem I got right by estimation. Since I'm watching from my phone in bed I needed a lazy solution. First x input is close to 0 and last x input is close to 1. F(0) = 1/3 and F(1) = 2/3. So the sum is a sum of a bunch of numbers in the range of 1/3 to 2/3 that are gradually increasing. Let's say the average value is the average of these fractions = 1/2. # of terms is 1996. Multiply by average value to get 998. I'm not claiming this is an accurate way to get the exact answer, but rather a good estimate. It just so happened in this case that my estimate was exactly right.
These are the types of solutions you're only going to remember when you're engaged with maths on a regular basis. Bravo.
Good problems.
I solved the first one in a school test in grade 9th, thats was easy if you observed the fact that f(x)+f(1-x)=1
The second one is Superr easy! I have done problem a couple of times.
Btw , Awespme explanation and animation❤❤❤❤
thank you.
I did the second problem slightly differently: If you take an initial step of putting everything in each parentheses over the same denominator then you get the total denominator as (2023!)^2, then pairing the + and - difference of squares terms gives a series of 1..2022 and 3..2024, which you can tweak to be expressed in terms of 2023! and the fraction falls out.
Moroccans love you !
By the way the series in the first problem works because 4 is the square of 2. If you change 4 and 2 to say 25 and 5 respectively, you would still get 998
So i would like to ask a website where i could get to see problem and their solution. I am preparing for some olympiad and i need some exercise to sharpen my mind
I solved it 😃
Similar question was asked in JEE ADVANCED 2020!!
In the 2nd problem, if you replace the problem with an infinite “series”, it converges to 1/2.
Yes. It's true 😊 (because the last term will tend to 1)
@@SidneiMVyes, I know!
In Korea, the above solution is learned at the age of 14~^^
4:34 Wow = and that's the answer.
Well yeah, since "and that's the answer" is said only in the end of each video.
I approximated the first problem using calculus: first let △=1/1997 and use riemann integral
f(△)△+f(2△)△+f(3△)△+...+f(1996△)△ ≈ ∫ f(x) dx with limits from 0 to 1
Divide both sides by △:
f(△)+f(2△)+f(3△)+...+f(1996△) ≈ (1/△) ∫ f(x) dx with limits from 0 to 1
Note that f(x) = (1/ln(4)) * ln(4)*4^x / (4^x+2) which can be integrated easily
f(△)+f(2△)+f(3△)+...+f(1996△) ≈ (1/ln(4)△) ln(4^x+2) with limits from 0 to 1
f(△)+f(2△)+f(3△)+...+f(1996△) ≈ (1/ln(4)△) (ln(6) - ln(3))
f(△)+f(2△)+f(3△)+...+f(1996△) ≈ 1997*ln(2)/ln(4)
f(△)+f(2△)+f(3△)+...+f(1996△) ≈ 1997/2
f(△)+f(2△)+f(3△)+...+f(1996△) ≈ 998.5
Re: Problem 1. The slightly more general function f(x) = a^x/(a^x+√a) for any a > 0 also satisfies f(x)+f(1-x) = 1. The only polynomial that satisfies it is f(x) = x. So, I'm curious what other
functions satisfy f(x)+f(1-x) = 1?
Problem 1 is a JEE (Main) Previous Year Question solved by high-school graduating students.
where did he get x998 from? 😢
the arithmetic mean of 2 approximations (1996/3 and 1996*2/3) i did for the first problem is 997.95, close enough for me xD
998
Can someone please explain to me how did we get 998
While you got that 1012/2023 is the answer for the second; you forgot to show that 1012 and 2023 are relatively prime. 2023 factors to 7 x 17 x 17. Since neither 7 or 17 is a factor of 1012; 1012/2023 is in lowest terms. Or you could have proven that a fraction in the form of n/(2n-1) is automatically in lowest terms.
That's a neat little point on the n/(2n-1). Let p > 1 be a factor of n (doesn't have to be prime but I'm using p anyway). Then n = pk for some integer k, so 2n - 1 = 2pk - 1. Clearly p is a factor of 2pk but is not a factor of 1, so p is not a factor of 2pk - 1. Thus, n and 2n - 1 have no common factors.
Those who know presh gave this q long ago
As an Algerian, We usually solve the problem from Morocco in less that 5 minutes at age of 17, not because we are Geniuses, we just see a lot of problems from this type.
well ours does it in less then 3 mins from same age
In my country we do these problems in about 5 seconds at age three, blindfolded while doing 100 press-ups
in my country we solve this problem in 5 days...
Well in Morocco, we solve this kind of problem at the age of 15 ( BTW I'm 15 )
Hi Presh, please read this feedback and implement it. I've said it many times and it doesn't seem to get anywhere. When you do a video with multiple problems, please either rearrange the videos so that you introduce the second problem after the first solution or at least review the second problem after the first solution. If I have to skip around in the videos in order to see it in a logical order, then the order must be wrong.
Hey dude, it's not that much of a big deal man it's only a maximum of 3 to 4 problems
Seemed pretty organized to me. Introduced the first question, then the second question, and then solved the first question, then the second question. If all else fails, Presh also took the liberty of labelling the parts of the video to be 1st solution and second solution.
I think the number of likes say most people agree with me. Thank you all for helping this feedback be so easily noticed! Myself, I usually do as much of the problem in my head as I can, so I'm not ready for the next problem until I've seen his solution for the current problem. Like I said, even if he doesn't change the order, at least give a refresher of the second problem before presenting its solution.
@mike1024. I'm glad that so many people agree with you. I am not saying you're wrong. I suppose it is only me that finds it easy to follow. I will glady help this get more attention if it helps you learn better in the future. Presh makes videos for people to be stimulated and learn, and if his current format isn't doing that for you, then perhaps he does need to change it. However, being the busy person they are, perhaps be grateful we have this content and make a step of your own to solve your dilema and use a pen and paper.
Why should he implement a change just to satisfy you ? Isn’t that a bit presumptuous?
1st problem has a mistake, it has an odd numer of elementes, the actual answer is 998.5
The numerators are counting the terms from 1 to 1996, which is an even number of terms. If this were, say 1/1998 to 1997/1998, I agree more work would have to be done with the middle term.
Presh Deadwalker
Every jee aspirants knows the ans .
I don't know why JEE aspirants think that they are something special.
JEE aspirants always show off in the comment section.