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Sum the Following Series: `Tan^-1 1/3+Tan^-1 2/9+Tan^-1 4/33+...+Tan^-1 (2^(N-1))/(1+2^(2n-1))` - Mathematics

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Question

Sum the following series:

`tan^-1  1/3+tan^-1  2/9+tan^-1  4/33+...+tan^-1  (2^(n-1))/(1+2^(2n-1))`

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Solution

`tan^-1  1/3+tan^-1  2/9+tan^-1  4/33+...+tan^-1  (2^(n-1))/(1+2^(2n-1))`

⇒ `tan^-1((2-1)/(1+2xx1))+tan^-1((4-2)/(1+4xx2))+tan^-1((8+4)/(1+8xx4))+...+tan^-1((2^n-2^n-1)/(1+2^n.2^(n-1))`

⇒ `(tan^-1  2-tan^-1  1)+(tan^-1  4-tan^-1  2)+(tan^-1  8-tan^-1  4)+...+(tan^-1  2^(n-1)-tan^-1 2^(n-2))+(tan^-1 2^n-tan^-1  2(n-1))`

⇒ `tan^-1 2^n-tan^-1  1`

⇒ `tan^-1 2^n -pi/4`

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Chapter 4: Inverse Trigonometric Functions - Exercise 4.11 [Page 82]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.11 | Q 4 | Page 82

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