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 evaluate a finite sum involving the inverse tangent function. The structure of the argument inside $\tan^{-1}$ suggests that this might be a telescoping series. After finding the sum, we can determine $\alpha$ and then $\tan \alpha$.
Step 2: Key Formula or Approach:
The key to simplifying this sum is the inverse tangent difference identity: \[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1} \left( \frac{x-y}{1+xy} \right) \] We will try to manipulate the general term of the series to fit this form.
Step 3: Detailed Explanation:
Let's analyze the general term inside the summation:
\[ T_p = \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right) \] We want to write the argument in the form $\frac{x-y}{1+xy}$.
Let's focus on the denominator's product term: $2^{2p-1} = 2^p \cdot 2^{p-1}$.
Let's try setting $x = 2^p$ and $y = 2^{p-1}$.
Now let's check if the numerator matches $x-y$:
$x-y = 2^p - 2^{p-1} = 2^{p-1}(2-1) = 2^{p-1}$. This matches the numerator perfectly.
So, we can rewrite the general term as a difference:
\[ T_p = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1}) \] Now, the summation becomes a telescoping series:
\[ \sum_{p=1}^{11} T_p = \sum_{p=1}^{11} [\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})] \] Let's expand the sum to see the cancellation:
\[ = (\tan^{-1}2^1 - \tan^{-1}2^0) + (\tan^{-1}2^2 - \tan^{-1}2^1) + \dots + (\tan^{-1}2^{11} - \tan^{-1}2^{10}) \] All intermediate terms cancel out, leaving the last term of the first part and the first term of the second part:
\[ \sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \tan^{-1}(2^0) = \tan^{-1}(2048) - \tan^{-1}(1) \] Since $\tan^{-1}(1) = \frac{\pi}{4}$, the sum is $\tan^{-1}(2048) - \frac{\pi}{4}$.
Now, we find $\alpha$:
\[ \alpha = \frac{\pi}{4} + \left( \tan^{-1}(2048) - \frac{\pi}{4} \right) = \tan^{-1}(2048) \] The question asks for $\tan \alpha$:
\[ \tan \alpha = \tan(\tan^{-1}(2048)) = 2048 \] Step 4: Final Answer:
The value of $\tan \alpha$ is 2048.