The answer is actually 2013*2^2013 according to my calculations, shifting the indices over by 1 as you copy the problem causes a slight difference in the solution. The method of solving the problem is the same, though.
That's because the problem shown on the first slide of the video is different from the problem on the second slide. I get 2013*2^2013 as well for the original problem
The best part about problems like this on proof-based tests is that once you see the solution, it’s typically almost trivia to actually prove that it’s correct.
First thing I tried was to calculate a_2 for which I have to set n = 1, then a_2 = (2/0)(a_1) = 4/0 which is undefined. So the rest of the sequence is undefined including a_2013 .
This problem appeared before in the British Maths Olympiad Round 1. The same method was used in the solutions to this paper in 'A Mathematical Olympiad Primer' by Geoff Smith.
yeay, this one i actually got, with pretty much the same method. (most of the time i try them but usually i end up having to watch the video to find out the solution. but i find if i try them at least a bit first i can both understand and appreciate the solution more.)
With this one in mind, it becomes easy to solve the following problem: (Source: 21st Philippine Mathematical Olympiad, Qualifying Round) A sequence {aₙ}ₙ ≥ ₁ of positive integers is defined by a₁ = 2, and for all integers n ≥ 2: 1/a₁ + 1/a₂ + 1/a₃ + ... + 1/aₙ ₋ ₁ + n/aₙ = 1. Find the numerical value of the sum of (k + 3)(3^k)/((4^k)a_k) from k = 1 to inf
You copied the problem over incorrectly, the problem you solved is different from the one defined in the actual problem, a_n=n/(n-2) * (a_1+...+a_n-1) but nice solution though
I have looked up the problem again. It should be written as a_n = (n+1/n-1)(a_1+...+a_n-1) for all n>= 2. There is a typo in the thumbnail, but the solution inside the video should be just fine.
@@letsthinkcritically really? It should be? It is intentionally and is a "modern" tactic to make viewers to ask themselves "how is it possible" and then to be forced to watch the video to see that it is not at all what was originally presented. Yesterday you had exactly the same "mistake" to a problem. These are not coincidence. Too bad, good at math, but ... as a human being. Unsubscribe.
Turned out to be easy, I got the answer a[2013]=2013*2²⁰¹¹. There must be a typo, sequence should start with a[2]=2, otherwise there's div by zero. Upd. Nope, the typo is different)) I just got previous term. Anyway, nice problem from the past.
In the thumbnail of the video I can see a_{n+1}=...; in the video there's written a_n=..., so correct the thumbnail, otherwise a_2 would be equal to [(1+1)/(1-1)]*..., which is nonsense.
Well for 14 year old and 15 year old kids it might seem a little intimidating but first but I also believe that it is a decent problem for a national mathematical Olympiad
It is not the first time you write a sentence and solve another sentence. If you do it only so that those who try can't solve it and have to watch your video, then you are showing a complete lack of seriousness. And how can you not see its pattern of an when a2 = 3 * 2 a3 = 4 * 4 a4 = 5 * 8 ... those 3, 4, 5... are from (n + 1) so it is impossible not to see the form of an. It's a simple sequence. The difficulty is that the pattern of an must be demonstrated by induction and it is not that simple. We start from (X ^ (n + 2) -1) / (X-1) = X ^ (n + 1) + ... + X + 1. Diferentiate left and right, replace X with 2, then divide by 2 and get the desired induction relation. Even your solution for the proof of an need's induction for completeness.
Concise and simple, very nice, youtube does need more quality math competition focused channels so keep doing what youre doing :)
Thank you very much!!
but those arent the same sequences at the start (maybe a small error) because the n+1 term was defined with (n+1/n-1) so the n term would be (n/n-2)
a_{n+1} = (n+1)/(n-1)*(a_1+...+a_n) or a_n=(n+1)/(n-1)*(a_1+...+a_{n-1}) ? I mean the first one does not make much sense for n=1.
The answer is actually 2013*2^2013 according to my calculations, shifting the indices over by 1 as you copy the problem causes a slight difference in the solution. The method of solving the problem is the same, though.
That's because the problem shown on the first slide of the video is different from the problem on the second slide. I get 2013*2^2013 as well for the original problem
The first problem would actually give an undefined value for a_2. The second problem fixes this mistake.
@@mullachv original problem case I get division by zero for a_2 (n=1)...
The best part about problems like this on proof-based tests is that once you see the solution, it’s typically almost trivia to actually prove that it’s correct.
First thing I tried was to calculate a_2 for which I have to set n = 1, then a_2 = (2/0)(a_1) = 4/0 which is undefined. So the rest of the sequence is undefined including a_2013 .
This problem appeared before in the British Maths Olympiad Round 1. The same method was used in the solutions to this paper in 'A Mathematical Olympiad Primer' by Geoff Smith.
Which year did it appear in BMO1?
letsthinkcritically sorry forgot to put the year! It was question 2 of the academic year 1996-1997
Amazing solution! Keep up the great work, and may I suggest some Brazilian Math Olympiad questions for later videos.
Thank you!!
Certainly, do let me know if there are any specific problems that you want me to talk about :)
Nice. Needed your help to get started but finished with the same answer!
I could sense a hint of Hong Kong accent. Are you from Hong Kong?
Yes, I am a HongKonger.
yeay, this one i actually got, with pretty much the same method. (most of the time i try them but usually i end up having to watch the video to find out the solution. but i find if i try them at least a bit first i can both understand and appreciate the solution more.)
great strategy!😌👏
Thank you!!
I did not understand from 5:30 to 6:20. How do you write (n+1)/2 ?
What application are you using for your discussion?
I use GoodNotes to write out the solutions
Please give solution for turkish junior national maths olympiad 2013 problem 2 as soon as possible
It is coming soon
Here you go! ruclips.net/video/oLGBpEV8GF0/видео.html
With this one in mind, it becomes easy to solve the following problem:
(Source: 21st Philippine Mathematical Olympiad, Qualifying Round)
A sequence {aₙ}ₙ ≥ ₁ of positive integers is defined by a₁ = 2, and for all integers n ≥ 2:
1/a₁ + 1/a₂ + 1/a₃ + ... + 1/aₙ ₋ ₁ + n/aₙ = 1.
Find the numerical value of the sum of (k + 3)(3^k)/((4^k)a_k) from k = 1 to inf
Sir plz upload some geometry problems also
a2 is = (2/0)*a1 is NOT defined. Does anyone have a reference to the actual problem?
Great video
Thank you very much!!
You copied the problem over incorrectly, the problem you solved is different from the one defined in the actual problem, a_n=n/(n-2) * (a_1+...+a_n-1) but nice solution though
I might have not got the indices exactly right but I thought they are referring to the same thing? It’s just the indices shifting by 1.
I have looked up the problem again. It should be written as a_n = (n+1/n-1)(a_1+...+a_n-1) for all n>= 2. There is a typo in the thumbnail, but the solution inside the video should be just fine.
@@letsthinkcritically really? It should be?
It is intentionally and is a "modern" tactic to make viewers to ask themselves "how is it possible" and then to be forced to watch the video to see that it is not at all what was originally presented.
Yesterday you had exactly the same "mistake" to a problem. These are not coincidence.
Too bad, good at math, but ... as a human being. Unsubscribe.
i like your video very much
Thank you!!
0:06 There is a mistake in the first slide, you have written the correct version on the next slide ig
amazing
Thank you!!
Turned out to be easy, I got the answer a[2013]=2013*2²⁰¹¹. There must be a typo, sequence should start with a[2]=2, otherwise there's div by zero. Upd. Nope, the typo is different)) I just got previous term. Anyway, nice problem from the past.
The typo was that [(n+1)/2]*2^(n-1)=(n+1)*2^(n-2), not (n+1)*2^(n-1)
In the thumbnail of the video I can see a_{n+1}=...; in the video there's written a_n=..., so correct the thumbnail, otherwise a_2 would be equal to [(1+1)/(1-1)]*..., which is nonsense.
BUT a2 is not defined.
so the trick at the end of the day is always remove the ugly parts : )
As a national Olympiad question, this is too simple
Well for 14 year old and 15 year old kids it might seem a little intimidating but first but I also believe that it is a decent problem for a national mathematical Olympiad
It is not the first time you write a sentence and solve another sentence. If you do it only so that those who try can't solve it and have to watch your video, then you are showing a complete lack of seriousness.
And how can you not see its pattern of an when
a2 = 3 * 2
a3 = 4 * 4
a4 = 5 * 8
...
those 3, 4, 5... are from (n + 1) so it is impossible not to see the form of an. It's a simple sequence.
The difficulty is that the pattern of an must be demonstrated by induction and it is not that simple.
We start from (X ^ (n + 2) -1) / (X-1) = X ^ (n + 1) + ... + X + 1. Diferentiate left and right, replace X with 2, then divide by 2 and get the desired induction relation.
Even your solution for the proof of an need's induction for completeness.
Calculation error in step 2. (n-2)*2 ≠ 2n-2.
That is actually n-2/n + 1 = n-2+n /n = 2n-2/n.