Question

If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) = \alpha$, then $\tan \alpha$ is equal to _________.

Answer 1

Correct Answer: 2048

Step 1: Understanding the Question:
We need to sum a series of inverse tangent functions and then find the tangent of the final expression.
Step 2: Key Formula or Approach:
Use the identity $\tan^{-1} \left( \frac{x-y}{1+xy} \right) = \tan^{-1} x - \tan^{-1} y$.
Step 3: Detailed Explanation:
Consider the general term of the sum:
\[ T_p = \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{p-1} \cdot 2^p} \right) \]
Notice that $2^p - 2^{p-1} = 2^{p-1}(2 - 1) = 2^{p-1}$.
So, $T_p = \tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \right) = \tan^{-1} (2^p) - \tan^{-1} (2^{p-1})$.
The sum is telescoping:
\[ \sum_{p=1}^{11} T_p = (\tan^{-1} 2^1 - \tan^{-1} 2^0) + (\tan^{-1} 2^2 - \tan^{-1} 2^1) + \dots + (\tan^{-1} 2^{11} - \tan^{-1} 2^{10}) \]
Sum $= \tan^{-1} (2^{11}) - \tan^{-1} (2^0) = \tan^{-1} (2048) - \tan^{-1} (1)$.
Sum $= \tan^{-1} (2048) - \frac{\pi}{4}$.
Given expression: $\alpha = \frac{\pi}{4} + (\tan^{-1} (2048) - \frac{\pi}{4}) = \tan^{-1} (2048)$.
Therefore, $\tan \alpha = 2048$.
Step 4: Final Answer:
The value of $\tan \alpha$ is 2048.