No worries, David. We are all lifelong learners. That's what makes our life exciting and meaningful! Thank you for your nice feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA!
Glad you enjoyed it! Thank you for your feedback! Cheers! You are awesome. Keep it up 😀 No wonder I reveal all kind of question for my audience that require various techniques... Love and prayers from the USA!
Dear Tim, if we scan this problem, we can see the powers of X and Y are incremented by 1 as we move downward. This gives us a good clue to multiply by (X+Y)! Moreover, these slick moves require consistent practice and continued exposure to such problems. No wonder I reveal all kind of question for my audience that require various techniques... Thanks for asking. Cheers! You are awesome. Keep it up 😀
Yes, it's May the 1st! So nice of you, Anatoliy Thank you for your wonderful feedback! Cheers! You are awesome. Keep it up 😀 Love and prayers from the USA! Peace!
@@pomlesty1348 it is difficult explain it here but I can tell you in short I just multiplied the 4th equation by (x+y) and I got ax⁵ and by⁵ with some other components which were deeply situated in other equation and that took me about 20-25 minutes. Hope it helped.
Let (1) t(n) = axⁿ + byⁿ then we have (2) t(n)·(x+y) = t(n+1) + xy·t(n−1) so (3) t(n+1) = (x+y)·t(n) − xy·t(n−1) Subtituting n = 2 and n = 3 in (3) and using the known values t(1) = 5, t(2) = 10, t(3) = 50, t(4) = 130 gives (4a) 50 = 10(x+y) − 5xy (4b) 130 = 50(x+y) − 10xy This is a linear system in x+y and xy with the solution x+y = 1, xy = −8. Substituting these values in (3) gives (5) t(n+1) = t(n) + 8·t(n−1) so we have a second-order linear homogeneous recurrence with constant coefficients. Substituting n = 4 in (5) and using t(4) = 130, t(3) = 50 gives t(5) = 130 + 8·50 = 530.
It came as something of surprise that neither {x,y} nor {a,b} were evaluated. It is possible to do so which makes for an interesting exercise. The result is : {x,y} = (1/2 + sqrt(33)/2, 1/2 - sqrt(33)/2 ), (1/2 - sqrt(33)/2, 1/2 + sqrt(33)/2) {a,b} = (1.6446,0.2303), (0.2303,1.6446)
Yes. As I explain in a different comment on this video, if we define (1) t(n) = axⁿ + byⁿ then t(n) satisfies (2) t(n+1) = t(n) + 8·t(n−1) so we have a second-order linear homogeneous recurrence with constant coefficients. To answer your question: we have (3) t(n+1) − t(n) = 8·t(n−1) In English: the difference between two consecutive values is equal to eight times the value preceding these two values.
Easy way is to generalized equation . Simple algerbra here is (ax^n + by^n) * ( x+y) = ax^n+1 + by^n+1 + xy (ax^n-1 + by^n-1) . This will make (2) * (x+y) = (3) + xy (1) and (3)* (x+y) =(4) +xy (2) . Then replace (2) with 10 , (3) with 50 , (4) with 130 . you will get x+y and xy of 1 and -8 . then put it to (4)*(x+y) = (5) + xy (3) . then answer is value of (5)
I write here for the first time. Concretely, we have solutions a = ( 165 - 75√33 )/528, b = ( 165 + 75√33 )/528, x = ( 1 - √33 )/2, y = ( 1 + √33 )/2. Of course, we can replace ( a, x ) with ( b, y ).
Petty point, but this always bugs me. The double distribution "rainbow method" is F.O.I.L firstxfirst outerxouter innerxinner lastxlast The word FOIL describes the complete process. Cheers
Yet another brilliant question (that left me totally bamboozled), followed by your customary elegant solution. Many thanks
No worries, David. We are all lifelong learners. That's what makes our life exciting and meaningful!
Thank you for your nice feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA!
Forgot to tell, sooo many steps explained with full passion and calm voice with full steps
I appreciate your tasks very much and it is a pleasure to follow them. Often I lack the ability to recognize certain patterns.
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
No wonder I reveal all kind of question for my audience that require various techniques...
Love and prayers from the USA!
This is higher than University level. Well done sir. I made A’s in University, but still find it hard.
The patience with which you explain each step is just amazing. Thank you Sir
So nice of you, Niru dear
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are the best. Keep it up 😀
Love and prayers from the USA!
@@PreMath 🙏🙏
@@nirupamasingh2948 hi again
@@srividhyamoorthy761 Continue with your hard work. You will excel
@@nirupamasingh2948 are you a college professor ?just asking
When I first saw the question
It became complicated
But you made it easy and simple
Thank you for your brilliant question and cool answers ❤❤
Fantastic.
It doesn't occur easily to bring in a multiplying factor ( x+y ) .
Great thinking..
sir , u are legend of mathematics.
So nice of you, Gupta dear
Thank you for your feedback! Cheers!
You are awesome. Keep it up. 👍
Love and prayers from the USA! 😀
Very good,I liked it very much.Thanks.
How did you come up with the idea of multiplying by (X + Y) in the first place?
My qsn also!
Dear Tim, if we scan this problem, we can see the powers of X and Y are incremented by 1 as we move downward. This gives us a good clue to multiply by (X+Y)! Moreover, these slick moves require consistent practice and continued exposure to such problems. No wonder I reveal all kind of question for my audience that require various techniques...
Thanks for asking. Cheers!
You are awesome. Keep it up 😀
I think x+y is a factor of the left expression
Nice and clearly explainrd, a crore subscribers cannot compensate for your lessons
Thank you for an interesting task, sir! My congratulation on the 1st May Holiday! God bless you and I wish all the best to you!
Yes, it's May the 1st!
So nice of you, Anatoliy
Thank you for your wonderful feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA! Peace!
I knew the tricky pattern all along.
Glad I was able to see it coming.
Finally after 30 minutes, I solved it my own but with different method
Btw thanks for this beautiful question ❤️❤️
Please share your method with us
@@pomlesty1348 it is difficult explain it here but I can tell you in short
I just multiplied the 4th equation by (x+y) and I got ax⁵ and by⁵ with some other components which were deeply situated in other equation and that took me about 20-25 minutes.
Hope it helped.
Well done!
You're welcome!
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA!
Thanks for video. Good luck sir!!!
You're welcome!
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA!
Well done. Three thumbs up!
Thanks
You're welcome!
Thank you for your feedback! Cheers!
You are awesome, Meng. Keep it up 😀
Love and prayers from the USA!
Hi, amazing video again! I love these videos but could it be possible to make these with a dark background instead of the white?
Let
(1) t(n) = axⁿ + byⁿ
then we have
(2) t(n)·(x+y) = t(n+1) + xy·t(n−1)
so
(3) t(n+1) = (x+y)·t(n) − xy·t(n−1)
Subtituting n = 2 and n = 3 in (3) and using the known values t(1) = 5, t(2) = 10, t(3) = 50, t(4) = 130 gives
(4a) 50 = 10(x+y) − 5xy
(4b) 130 = 50(x+y) − 10xy
This is a linear system in x+y and xy with the solution x+y = 1, xy = −8. Substituting these values in (3) gives
(5) t(n+1) = t(n) + 8·t(n−1)
so we have a second-order linear homogeneous recurrence with constant coefficients. Substituting n = 4 in (5) and using t(4) = 130, t(3) = 50 gives t(5) = 130 + 8·50 = 530.
Great, thank you Sir .
Sir are you a great teacher 👩🏫
So nice of you
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA!
Ottimo professore
the only hard thing is how to evaluate that factor x+y will determine the solution, how to find out that?
wow nice the perfect solution
Bravo! Frumos! Respect! 👏🤗🥇👀🇦🇩
Very interesting.
Excellent
So nice of you, Sanjay
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Brilliant
u r awesome! keep it up you r the best
Tysm :')
It came as something of surprise that neither {x,y} nor {a,b} were evaluated.
It is possible to do so which makes for an interesting exercise.
The result is :
{x,y} = (1/2 + sqrt(33)/2, 1/2 - sqrt(33)/2 ), (1/2 - sqrt(33)/2, 1/2 + sqrt(33)/2)
{a,b} = (1.6446,0.2303), (0.2303,1.6446)
¡Fantástico!
this was very well explained, thanks so much for sharing
Do the differences between the equations (5,40,80,400) mean anything?
Yes. As I explain in a different comment on this video, if we define
(1) t(n) = axⁿ + byⁿ
then t(n) satisfies
(2) t(n+1) = t(n) + 8·t(n−1)
so we have a second-order linear homogeneous recurrence with constant coefficients. To answer your question: we have
(3) t(n+1) − t(n) = 8·t(n−1)
In English: the difference between two consecutive values is equal to eight times the value preceding these two values.
awesome!
Very nice explanation 🙂
Thnku
You're welcome!
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA!
Əla həll etdiniz.Təşəkkürlər.Bakıdan salamlar.
Done👍
First view
Fantastic!
Thank you for your feedback! Cheers!
You are awesome, Goel. Keep it up 😀
Very nice 👍
Glad you enjoyed it!
Thank you for your feedback! Cheers!
You are awesome, Chan. Keep it up 😀
BRO, GOOD pRESENTATION. wELLCOME.
Easy way is to generalized equation . Simple algerbra here is (ax^n + by^n) * ( x+y) = ax^n+1 + by^n+1 + xy (ax^n-1 + by^n-1) . This will make (2) * (x+y) = (3) + xy (1) and (3)* (x+y) =(4) +xy (2) . Then replace (2) with 10 , (3) with 50 , (4) with 130 . you will get x+y and xy of 1 and -8 . then put it to (4)*(x+y) = (5) + xy (3) . then answer is value of (5)
Bravo!!! (Thanks!)
It's genius
First we need to prove that there exist a,b,x,y that would satisfy given equations. Then calculations are correct.
asnwer=60 my try hard study
احسنت
لطف منك يا أخي العزيز محمد.
شكرا لك على ملاحظاتك! هتافات!
انت رائع 😀
حب ودعاء من امريكا!
I write here for the first time. Concretely, we have solutions
a = ( 165 - 75√33 )/528,
b = ( 165 + 75√33 )/528,
x = ( 1 - √33 )/2,
y = ( 1 + √33 )/2.
Of course, we can replace ( a, x ) with ( b, y ).
Super job!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
Love and prayers from the USA!
Petty point, but this always bugs me.
The double distribution "rainbow method" is F.O.I.L firstxfirst outerxouter innerxinner lastxlast
The word FOIL describes the complete process.
Cheers
Interesting mathematical gymnastics. How is this applied practically?
❤️
步驟拉太多，變數變換有夠多餘，只需要解出xy=-8就可以了
👍👍👍🙏🙏🙏👏👏👏
Do we need that in our life? I believe that we need to teacher real situation to the alumni….🤷🏻♂️
thanks