You have pleased me .Your method of proving the problem is extraordinary ,dear professor .Keep it up !! May God bless you!!!
@themathsgeek8528
2 жыл бұрын
This was a nice problem with a great solution! Thanks for the video
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@Junieper
2 жыл бұрын
Show that the alternating series m^n-(n choose 2)*(m-1)^n)+ (n choose 3)* (m-2)^n… =0 when m>n
@fierydino9402
2 жыл бұрын
Thank you for sharing such elegant problem and solution 😀
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@MrLidless
2 жыл бұрын
Beautiful.
@user-vg1qo5gi3l
2 жыл бұрын
Nice! Really interesting trick
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@lalitmarwaha3537
2 жыл бұрын
This was a very enjoyable question i got the feeling of it but i couldn't manipulate accordingly
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@DaveyJonesLocka
2 жыл бұрын
This is a nice trick to add to anyone’s problem-solving arsenal: recognizing a ratio of products, and rewriting it as some sort of combination.
@txikitofandango
2 жыл бұрын
The way I see it, the nth term is 2(2n+1)/n; therefore, the product equals 2ⁿ(2n-1)!!/n!. If n is odd, and assuming that the product of the first n-1 terms is an integer, then the new denominator will be cancelled out in a previous numerator n = 2k-1 for some k. If n is even, then n =2ᵏ(2m-1). The odd factor will be cancelled out as already mentioned. As for the power of 2 in the denominator, you can show that n! contains at most n-1 factors of 2. So that accounts for all factors in the denominator. EDIT: And that's pretty much how you did it! Great minds! In fact, your way with combinatorics is very neat. Great idea!
@eccleshillstluke5352
2 жыл бұрын
When I got to the end of your video I suddenly realized that if the product of all these terms really is (2n Choose n), then what you have proved is: (2n Choose n) = (4 - 2/n) times ((2n-2) Choose (n-1)), which indeed it is, as the following reasoning shows: "In order to choose n items from 2n, first choose one of them (in 2n ways), and then choose one of the remaining 2n-1 to reject (in 2n-1 ways). Then choose n-1 items from the remaining 2n-2. Each final result has been arrived at in n^2 ways (because the initial choice and the initial reject were each one of n possibilities from the final result), so the multiplicand to get from (2n-2 Choose n-1) to (2n Choose n) is 2n(2n-1)/n^2 = (4n-2)/n = 4-2/n." So by induction (starting with (2 Choose 1) = 2 = 4-2), the product of the first n terms of the sequence of brackets is 2n Choose n, which as you rightly observe is an integer.
@willbishop1355
2 жыл бұрын
I tried to use an induction proof to show that if (4 - 2/1)*...*(4 - 2/n) is divisible by n+1, then the same product times 4 - 2/(n+1) is divisible by n+2. This seems to be the reason we always get an integer (each product is divisible by the next integer), but I couldn't get it to work.
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@petersievert6830
2 жыл бұрын
The problem is, that it is not enough. E.g. if you had purely theorically 8 for n=3, thus divisible by 4, your next number would be 28, which is in turn not divisible by 5. You would need a stronger induction hypothesis after all.
Any way to show that (2n)!/(n!)^2 is an integer without knowing it is the same as 4C2 or anything about combinatorics?
@johnchessant3012
2 жыл бұрын
The number of factors of p in (2n)! is [2n/p] + [2n/p^2] + [2n/p^3] + ... (brackets are floor function) The number of factors of p in (n!)^2 is 2[n/p] + 2[n/p^2] + 2[n/p^3] + ... It's clear that [2n/p] >= 2[n/p], etc. so (2n)! has more factors of p than (n!)^2, for all p.
@SpeedyMemes
2 жыл бұрын
pick a test case, e.g. n = 5, write out the top and bottom in full, and see if you notice anything
@HeavyMetalMouse
2 жыл бұрын
@@johnchessant3012 That's a pretty elegant way to do it. I like that. :)
@byronrobbins8834
2 жыл бұрын
@@johnchessant3012 well, however 2*3 = 6, then we get the number 20, and then 20 * 3.5 = 70, 70 * 3.6 = 252, and so on, like as if Sowon gets the pants and jacket, and Umji ends up in a Kilt.
@user-jc4kc9od3n
2 жыл бұрын
@@byronrobbins8834 ugguug7g
@jfcrow1
2 жыл бұрын
so approaches 4^n for 2n choose n
@georgesbv1
2 жыл бұрын
Could have divided back the 2n! to show that it is integer
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@yoav613
2 жыл бұрын
This is (2n,n),very simple to anyone who learned gamma function and its properties like gamma(n+1/2).
@cobokobo2115
2 жыл бұрын
amazing...kzitem.info/news/bejne/z4h606ugfZ57m6g
@user-of9sr8bm9i
2 жыл бұрын
You just saying what he did
@sithlordbinks
2 жыл бұрын
Way to ruin the fun
@bekkaouimohammed1808
Жыл бұрын
Gréât
@djvalentedochp
2 жыл бұрын
I went to wikipedia to figure out some relations involving double factorials and then I was able to solve this problem making use of other methods The product above can be reduced to 2^n (2n - 1)!! / n! Then, we can apply the following properties to simplify even more: (2k)!! = k! 2^k (2n - 1)!! = (2n - 1)!/(2n - 2)!! = (2n - 1)! / [(n - 1)! 2^(n - 1)] Applying all that to the original expression and we get a binomial coefficient, which is clearly an integer
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