JEE Delight | Sequence & Series | Sophie Germain Identity | A rare telescoping product problem

JEE Delight | Sequence & Series | Sophie Germain Identity | An elegant solution to a rare telescoping product problem.

Solve 1^4+1/4)(3^4+1/4)...((2n-1)^4+1/4))/(2^4+1/4)(4^4+1/4)...((2n)^4+1/4))

Unravel the secrets of a challenging JEE Advanced math problem with our succinct and insightful solution. If you thought higher algebra was daunting, think again! Our video breaks down a complex sequence problem, showing you a step-by-step approach to finding the elusive values of a, b, and c. Perfect for JEE aspirants looking to master their problem-solving skills and gain that competitive edge. Watch, learn, and conquer your math fears as you prepare for one of the most pivotal exams of your life. Join us on this mathematical adventure and stay ahead of the curve with more tips, tricks, and solutions!

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[aquafinagaming]

SIR YOUR CHANNEL IS REALLY AWESOME MAY BE IT WILL TAKE SOME TIME BUT YOU WILL REALLY HAVE A HUGEE AMOUNT OF STUDENTS FOLLOWING YOU BCOZ U R FILLING THE GAP OF MATHEMATICS THAT IS LEFT BEHIND IN JEE PREPARATION

[Your_Study_Buddy_SD]

Sir as you know, quite often questions in maths in jee advanced are asked in such a way that solving them by their normal methods is quite long. Whereas there may be a certain theorem or formula through which that question can be solved in just a couple of steps.
How can we practice on such theorems and formulas? Is there any source for them?

[harshuldesai]

Substitute 3 different values for n. Solve the system for a, b and c.

[aseemaggarwal]

Sir please merge conic sections 🙏🙏. Also sir by completing your 4hr theory vid on pnc + all prac videos . Can we say our pnc is done wrt concepts ?

[VJustice]

Keep it going sir you're doing great

[CosmosDev-qs]

Sir, can you please provide sources for such level of questions. I'll be more than grateful 🙏

[SrirupNandi]

This is question from yellow book. First time i tried, couldn't move it at all

[advarbindkumar]

Sir pls make a video on fermat's last Theorem

[scpforjee]

Wow I had the exact same question in a test yesterday.

[Rahul-brsz]

Can be done with sophie german identity

[honestadministrator]

f(r) = r^4 + 1/4 = (r^2 + 1/2) ^2 - r^2
= (r^2 + 1/2 + r) (r^2 + 1/2 - r)
f(r + 1)
= (r^2 + 3 r + 5/2) (r^2 + r + 1/2)
f(r) /f ( r + 1)
= (r^2 + 1/2 - r)/ r^2 + 3 r + 5/2)
= ( r^2 + 1/2 - r)
/ ( ( r + 2) ^2 + 1/2 - ( r +2))
= g (r) /g ( r +2), say
Hereby
f(1) f (3).. f (2 n - 1)
/ f(2) f (4).. f (2 n )
= [g(1) / g(3) ]
* [g(3) / g(5) ]
..
** [g(2n -1) / g(2 n + 1) ]
= g(1) /g ( 2 n + 1)
= (1^2 + 1/2 - 1)
/( ( 2 n + 1) ^2 + 1/2 - ( 2 n + 1))
Hereby
a n^2 + b n + c
=[ (2 n + 1) ( 2 n) + 1/2 ] / (1/2)
= 4 n ( 2 n + 1) + 1
= 8 n^2 + 4 n + 1
Hereby a - b + c = 8 - 4 + 1 = 5