Question

If \( \alpha = \frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) \), then the value of \( \tan(\alpha) \) is:

• A. \( 2^9 \)
• B. \( 2^{10} \)
• C. \( 2^{11} \)
• D. \( 2^{12} \)

Hint:

In \( \tan^{-1} \) summations, always try to split the numerator into a difference of two factors whose product (plus 1) matches the denominator.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The summation involving \( \tan^{-1} \) can be solved by expressing the argument as a difference of two terms, \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right) \), leading to a telescoping series.

Step 2: Key Formula or Approach:
Rewrite the general term: \( \frac{2^{p-1}}{1 + 2^{2p-1}} = \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \).

Step 3: Detailed Explanation:
1. General term \( T_p = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1}) \). 2. Sum \( S = \sum_{p=1}^{11} [\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})] \). This is a telescoping sum: \( S = (\tan^{-1} 2^1 - \tan^{-1} 2^0) + (\tan^{-1} 2^2 - \tan^{-1} 2^1) + \dots + (\tan^{-1} 2^{11} - \tan^{-1} 2^{10}) \). \( S = \tan^{-1}(2^{11}) - \tan^{-1}(1) = \tan^{-1}(2^{11}) - \frac{\pi}{4} \). 3. Substitute back into \( \alpha \): \( \alpha = \frac{\pi}{4} + \left( \tan^{-1}(2^{11}) - \frac{\pi}{4} \right) = \tan^{-1}(2^{11}) \). 4. Therefore, \( \tan(\alpha) = 2^{11} \).

Step 4: Final Answer:
The value of \( \tan(\alpha) \) is \( 2^{11} \).