1. Tardigrade
  2. Question
  3. Mathematics
  4. A telescoping series is a series whose general term tn can be written as tn=an-an-1 i.e the difference of two consecutive terms of a sequence (an). The value of ( sec x/2)+( sec x sec 2 x/22)+( sec x sec 2 x sec 22 x/23)+ ldots ldots 8 terms = cos θ(0< θ< (π/2)) when x=(π/512) then cos (1024 θ)=

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Q. A telescoping series is a series whose general term $t_n$ can be written as $t_n=a_n-a_{n-1}$ i.e the difference of two consecutive terms of a sequence $\left(a_n\right)$.
The value of $\frac{\sec x}{2}+\frac{\sec x \sec 2 x}{2^2}+\frac{\sec x \sec 2 x \sec 2^2 x}{2^3}+\ldots \ldots 8$ terms $=\cos \theta\left(0< \theta< \frac{\pi}{2}\right)$ when $x=\frac{\pi}{512}$ then $\cos (1024 \theta)=$

JEE AdvancedJEE Advanced 2021

Solution:

Correct answer is '1'