The fact that sqrt(1+1/n^2+1/(n+1)^2) is equal to 1+1/n-1/(n+1) totally blew my mind. That minus there feels like magick. And you just breezed through it as if it is obvious.
I got to the point where I knew each term was 1+1/n+1/(n+1) far more iteratively. Just doing the math, the first term is sqrt(9/4)=3/2. The second is sqrt(49/36)=7/6. The third is sqrt(169/144)=13/12. Examining each of these for patterns, with emphasis on terms in the form of 1/x and 1/(x+1) (since those were involved in the original problem) quickly reveals that each term is 1, plus the difference between 1/n and 1/(n+1). At that point, jump to 6:15 of the video, and finish the same way.
@@Bob94390 I Never Said it was but it is great tool for getting a basic understanding, and patterns can be wrong as well( an example being Moser's circle proble)
These types of problems always begins at the beginning starting with stress and high Temper and ends with relieve when it is generalized to the maximum extent🙂🙃
Adjusting the numerator mathematically as perfect square is bit tough job. But observing the first few terms like 2*1+1=3 in first term . In second term it's 3*2+1=7 etc .by induction we conclude generalise numerator is (n+1)*n+1 .
Testing two terms shows that every term is likely rational 1 +1/n^2 +1/(n+1)^2 = 1 + (2n^2+2n+1)/(n(n+1))^2 = 1 + 2/n(n+1) + 1/(n(n+1))^2 = (1+1/n(n+1)) So every term is 1 + 1/n(n+1) 1/n(n+1) is a famous telescopic sum series ans: 2007 + 1/1 - 1/2008 = 2007 + 2007/2008
is it possible to solve this problem without telescoping? I tried looking at the pattern of how each value in the sum changes and found that for every value from the starting value 3/2 is equal to 3/2 with 4+2(n-1) [changing n to 2006] added to the numerator and denominator. I then tried rewriting the equation using this pattern and then solve for the value of the sum. the value I get in my calculator is 2174.791 which is off from the 2008-(2007/2008). maybe you can't use the finite sum equation for this type of equation or maybe I did my math wrong; probably both. anyone know?
sqrt(4) = 2 and not -2 However the solution to the equation x^2 = 4 is x1 = sqrt(4) = 2 and x2 = -sqrt(4) = -4 The square root function, just like any function, only gives one image to an argument, and for the square root function it is the positive number
@@user-yt198 Well not an integer perfect square but it makes it a ratio fraction on each one, yes, that's the part I am saying is crazy. Do you have a problem with it?
Ok, it's 2230 & one look at this thumb & I know...tonight I'm a passenger on today's Presh Express! My nephew watched the Squid Game & has subbed. Now to start the video, no paper or calc.. 🇦🇺
Nice problem. I saw that it reduced to Ʃ √(n⁴ + 2n³ + 3n² + 2n +1)/(n(n + 1)) The symmetry in the numerator led me to u = n + 1/n It reduces quickly to Ʃ (u + 1)/(n + 1) = Ʃ (1 + 1/(n(n + 1))) = 2007 + Ʃ (1/n - 1/(n+1)) = 2007 + 1 - 1/2008 = 2008 - 1/2008
![Blox Fruits Dragon Rework Update [Full Stream]](http://i.ytimg.com/vi/EqDAp8udhm0/mqdefault.jpg)
Using 2008 as max value in a 2020 contest. Weird. Would work with 2025 as well.
I don't understand why he didn't actualised the problem, we are in january 2025 that was the perfect time
If they were to make it 2048 it would be pure evil, throwing contestants intuition down a completely wrong path.
I see sum,I see Ramanujan,I answer -1/12.
The thruth is that the answer is epsilon(-1/12), where epsilon is a scuffed function
I think I'm wrong
@Stagnated541 Nice!
LMAO😂
Bro it's brahmahgupta not ramanujan
@mostafabenzarti6205 Its Ramanujan
The fact that sqrt(1+1/n^2+1/(n+1)^2) is equal to 1+1/n-1/(n+1) totally blew my mind. That minus there feels like magick. And you just breezed through it as if it is obvious.
You could also express the answer as (2007/2008)*2009.
n+1-1/(n+1)=((n+1)^2-1)/(n+1)=(n*n+2*n)/(n+1)=n*(n+2)/(n+1)
I got to the point where I knew each term was 1+1/n+1/(n+1) far more iteratively. Just doing the math, the first term is sqrt(9/4)=3/2. The second is sqrt(49/36)=7/6. The third is sqrt(169/144)=13/12. Examining each of these for patterns, with emphasis on terms in the form of 1/x and 1/(x+1) (since those were involved in the original problem) quickly reveals that each term is 1, plus the difference between 1/n and 1/(n+1).
At that point, jump to 6:15 of the video, and finish the same way.
pretty much same as I did just pattern rec
Guessing based on perceived patterns is not a perfect replacement for a rigorous proof.
@@Bob94390 I Never Said it was but it is great tool for getting a basic understanding, and patterns can be wrong as well( an example being Moser's circle proble)
The extent of how this question simplified itself is just magic
The answer can also be expressed as 4,016,063/2008.
These type of questions are still existing 😢😢😢..Thanks for sharing
These types of problems always begins at the beginning starting with stress and high Temper and ends with relieve when it is generalized to the maximum extent🙂🙃
I guessed about 2007 and I am ok with it
Ummm looks like telescopic sun, pretty cool, lets try
Brilliant question with brilliant solution 🎉
(sum sqrt(1+1/k^2+1/(k+1)^2) k=1 to n)=n+1-1/(n+1)
Adjusting the numerator mathematically as perfect square is bit tough job. But observing the first few terms like 2*1+1=3 in first term . In second term it's 3*2+1=7 etc .by induction we conclude generalise numerator is (n+1)*n+1 .
Wow, that's amazing!
Did I fall asleep for 17 years?
Yes. Let me explain. No, it would take too long. Let me sum up.
2025 - 1/2025
@fluffysheap Inconceivable.
it would have been better if you showed the expreession for any n, although it is quite obvious with the result
I found the answer to be 2008-(1/2008). Feels nice to get this one. Wasn't too much trickery involved
Testing two terms shows that every term is likely rational
1 +1/n^2 +1/(n+1)^2 = 1 + (2n^2+2n+1)/(n(n+1))^2 = 1 + 2/n(n+1) + 1/(n(n+1))^2 = (1+1/n(n+1))
So every term is 1 + 1/n(n+1)
1/n(n+1) is a famous telescopic sum series
ans: 2007 + 1/1 - 1/2008 = 2007 + 2007/2008
Amazing question ❤❤
Next level question sir🙏🏼
i solved it in two tries.
Lovely problem! I just managed that on in my head but it was nice and challenging.
You can also write the answer as 4032063/2008
is it possible to solve this problem without telescoping? I tried looking at the pattern of how each value in the sum changes and found that for every value from the starting value 3/2 is equal to 3/2 with 4+2(n-1) [changing n to 2006] added to the numerator and denominator. I then tried rewriting the equation using this pattern and then solve for the value of the sum. the value I get in my calculator is 2174.791 which is off from the 2008-(2007/2008). maybe you can't use the finite sum equation for this type of equation or maybe I did my math wrong; probably both. anyone know?
before watching the video i thought the answer is 2008 plus something positive near zero, and I surprised with the real answer
This math problem must have rocked in 2007.
Mind my ignorance. What about the negative solutions of the square roots? How can they be ruled out without even mentioning?
sqrt(4) = 2 and not -2
However the solution to the equation x^2 = 4 is x1 = sqrt(4) = 2 and x2 = -sqrt(4) = -4
The square root function, just like any function, only gives one image to an argument, and for the square root function it is the positive number
@ isn’t the sqrt(x) graph a parabola laid horizontally? For every x there are two values of sqrt(x).
@@jfg31416 Not when you view square root as a single-valued function, which is normally assumed unless context dictates otherwise.
@@chrischappa962 that is precisely my question. Why is this assumption correct in this case?
@@jfg31416
Because the radical itself indicates only the positive root.
Generalize by induction?
if you dont want to see a lot of n in this process you can write first three process and the last process , you will see the pattern
Crazy how none of those 2007 square roots are irrational.
🤨
@@jordan7828 What's wrong?
Because each of them has a sum of a perfect square?
@@user-yt198 Well not an integer perfect square but it makes it a ratio fraction on each one, yes, that's the part I am saying is crazy. Do you have a problem with it?
Thought it would be pi x 1000..
Ok, it's 2230 & one look at this thumb & I know...tonight I'm a passenger on today's Presh Express! My nephew watched the Squid Game & has subbed. Now to start the video, no paper or calc.. 🇦🇺
Can’t u just say sqrt(1+x)=1+x/2… then (1/n^2+1/(n+1)^2))/2 is the mean which equals 1/(n(n+1))… so u get 1+1/(n(n+1)) immediately???
It was doable gotta say
Nice one
This one came in my reshuffling exam of fiitjee coimbatore, 2022.
Why do you no longer tell the viewers to pause the video after you've explained the problem to give them a shot at solving it themselves?
did the music get copyright striked?
I thought the answer was 2008.
Sad I was off by - 0.0005
Old classic!
Thanks
Nice problem. I saw that it reduced to Ʃ √(n⁴ + 2n³ + 3n² + 2n +1)/(n(n + 1))
The symmetry in the numerator led me to u = n + 1/n
It reduces quickly to Ʃ (u + 1)/(n + 1) = Ʃ (1 + 1/(n(n + 1))) = 2007 + Ʃ (1/n - 1/(n+1)) = 2007 + 1 - 1/2008 = 2008 - 1/2008
Or… instead of doing all that fancy factoring.. just look at the problem!❤️😅 What does sqrt(1/(n^2)) always equal!!
1/2008=0.000498
2008-0.000498=2007.999502
Very nice👍
Mind blowing
Very cool.
My calculation led to 2007+1/2006.... well close enough😀
Nice!!!
Justawsman!
I would just integrate
Just trying to do this in my head, but I think the answer is… [SPOILER]
2008 - 1/2008?
Yay im the first
#free_gaza
No, no freeing Gaza. Your political propaganda does not belong here. *Get lost!*
#free_palestine
Your political propaganda does not belong here. *Get lost!*
something around 2008 ???
_Answer_ : (2008 - 1/2008) = 4032063/2008
_Calculation_ :
1 + 1/n² + 1/(n+1)² =
= (n²(n+1)² + (n+1)² + n²) / [n²(n+1)²]
= ( (n²+1)(n+1)² + n²) / [n²(n+1)²]
= ( (n²+1)(n² + 2n +1) + n²) / [n²(n+1)²]
= ( n²*(n² + 2n +1) + (n² + 2n +1) + n²) / [n²(n+1)²]
= ( n²*n² + n²*2n + n² + n² + 2n +1 + n²) / [n²(n+1)²]
= ( n⁴ + 2n³ + 3n² + 2n + 1) / [n²(n+1)²]
= ( (n²+n+1)² ) / [n²(n+1)²]
= ( (n²+n+1) / [n(n+1)] )²
= [(n²+n+1)/(n²+n)]²
Hence,
√(1 + 1/1² + 1/2²) + √(1 + 1/2² + 1/3²) + ... + √(1 + 1/2007² + 1/2008²) =
= Σ √( 1 + 1/k² + 1/(k+1)² ) , from k=1 to k=2007
= Σ √( [(k²+k+1)/(k²+k)]² ) , from k=1 to k=2007
= Σ (k²+k+1)/(k²+k) , from k=1 to k=2007
= Σ (1 + 1/(k²+k)) , from k=1 to k=2007
= { Σ (1) , from k=1 to k=2007 } + { Σ 1/(k²+k) , from k=1 to k=2007 }
... note: first summation equals 2007 , in second summation 1/(k²+k) = 1/k - 1/(k+1) ...
= 2007 + { Σ (1/k - 1/(k+1)) , from k=1 to k=2007 }
= 2007 + { Σ 1/k , from k=1 to k=2007 } - { Σ 1/(k+1) , from k=1 to k=2007 }
... in the last summation, substitute j = k+1 ...
= 2007 + { Σ 1/k , from k=1 to k=2007 } - { Σ 1/j , from j=2 to j=2008 }
= 2007 + 1/1 + { Σ 1/k , from k=2 to k=2007 } - { Σ 1/j , from j=2 to j=2007 } - 1/2008
= 2007 + 1 - 1/2008
= 2008 - 1/2008
= 2007 + 2007/2008
= (2008² - 1)/2008
= (2007*2009)/2008
= (4000000 + 32000 + 64 - 1)/2008
= (4032063)/2008