An Infinite Radical With A Special Flavor

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
/ @aplusbi
❤️ ❤️ ❤️ My Amazon Store: www.amazon.com...
When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you.
If you are preparing for Math Competitions and Math Olympiads, then this is the page for you!
CHECK IT OUT!!! ❤️ ❤️ ❤️
❤️ A Differential Equation | The Result Will Surprise You! • A Differential Equatio...
❤️ Crux Mathematicorum: cms.math.ca/pu...
❤️ A Problem From ARML-NYSML Math Contests: • A Problem From ARML-NY...
❤️ LOGARITHMIC/RADICAL EQUATION: • LOGARITHMIC/RADICAL EQ...
❤️ Finding cos36·cos72 | A Nice Trick: • Finding cos36·cos72 | ...
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries
#algebra #numbertheory #geometry #calculus #counting #mathcontests #mathcompetitions
via @KZbin @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus

[@ЕленаСибирячка-к5у]

It's fantastic!

[@RobG1729]

Nice accompnying my morning coffee!

[@FisicTrapella]

very nice!! 👌

[@Blaqjaqshellaq]

That last sum isn't e - (e-1), it's (e-1) - (e-2).

[@maxvangulik1988]

i just turned it into a telescoping series. I wasn't evaluating allat.

[@snejpu2508]

You can also calculate the value of the series by taking the series for e^x, dividing both sides by x, taking the derivative of both sides, and substitution x=1. It gives you the answer of 1. :-)

[@SyberMath]

Don't we get an extra -1 at the end if I did it correctly?

[@black_eagle]

Yay I got this one. Took me a while to get the expression in terms of a product of exponents of 2, I wasn't clever enough to see that I could just distribute the exponents inside the radicals so I came up with a recurrence relation using ratios of successive terms and eventually found the pattern. A nice contrived problem ;)

[@scottleung9587]

Awesome!

[@sadeghradjai7716]

Bravo...very nice

[@gregevgeni1864]

Smart 🎉🎉

[@maxwellarregui814]

Señores: SyberMath. Reciban un cordial saludo, Gracias por este ejercicio muy bueno, ya logré entenderlo; por favor sería bueno otro parecido como refuerzo. Éxitos.

[@SyberMath]

¡Gracias!

[@positivenozy6065]

Interesting.. Thank you for sharing this problem. Could you tell me how to solve this kind of calculus tasks? Hard level or even Olympiad. Looking forward for your answer!

[@leonardobarrera2816]

Power series?? Hahaha

[@phill3986]

👍Nice, but for good form I'd show limit of last term goes to zero as n goes to infinity

[@LC95297]

Limit when n tends to infinite of 2^[n/(n+1)], which is 2¹ so 2.

[@buff9943]

I found 2**(3/2)

[@nasrullahhusnan2289]

Let x=sqrt[2×cbrt{4×qtrt(8...)}] =sqrt[2×cbrt}2²×qtrt(2³...)}] =sqrt[cbrt{2⁵×qtrt(2³...)}] =sqrt[cbrt{qtrt(2²³...)}] =(2²³...)^(½×⅓×¼×...) Therefore x=2

[@usmonaliumarov1136]

Iʼm from Uzbekistan Doim kuzataman

[@yusufdenli9363]

Time 7.57 That is not e. That is (e -1) And the other one is (e - 2) (e -1) - (e - 2) =1

[@SyberMath]

oopsies!

[@misterdubity3073]

Quick and dirty estimate. Pretend there is no (...). Go step by step starting with the 8, then the 4th root, then the 4 etc: 2^3; 2^3/4; 2^11/4; 2^11/12; 2^23/12; 2^23/24 looks pretty close to 2. Faster way maybe for a multiple choice test.

[@Nothingx303]

It's Πορτοκαλί flavor 😋 😜 🤪

[@armacham]

another way to solve: define a function f(n) such that: f(n) = (2^(n-1) * f(n+1) ) ^(1/n) and we want to find f(2) logbase2 both sides to get: log2(f(n)) = (1/n) * log2( 2^(n-1) * f(n+1) ) log2(f(n)) = (1/n) * ( n - 1 + log2(f(n+1)) log2(f(n)) = 1 - 1/n + log2(f(n+1))/n log2(f(n)) - 1 = (log2(f(n+1)) - 1)/n define a new function g(n) = log2(f(n)) - 1 now we can say g(n) = g(n+1)/n so g2 = g3/2 and g3 = g4/3 and g4 = g5/4 if g2 = g3/2 then we can substitute in g4/3 for g3 to get: g2 = g4/6 and from here it's clear g2 = g(m+1)/m! for all values of m we can show that g(m+1) increases slower than m! increases so that as m goes to infinity, g(m+1)/m! = 0 meaning g(2) = 0 use the definition of g(2) to get: g(2) = log2(f(2)) - 1 = 0 log2(f(2)) = 1 f(2) = 2 and that's the answer we were looking for

[@maxvangulik1988]

R=2^1/2•4^1/6•8^1/24•... R=prod[n=2,♾️]((2^(n-1))^(1/n!)) R=prod[n=2,♾️](2^((n-1)/n!)) ln(R)=ln(2)•sum[n=2,♾️]((n-1)/n!) ln(R)=ln(2)•sum[n=2,♾️](1/(n-1)!-1/n!) ln(R)=ln(2) R=2

[@yoav613]

Noice!