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

The key to solving this problem is to use the Method of Differences (Telescoping Series) by applying the identity:

$$\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$$

The general term in the summation is $$T_p = \tan^{-1} \left( \frac{2^{p-1}}{1 + 2^{2p-1}} \right)$$.

We can rewrite the numerator and denominator to fit the identity:

• Numerator: $$2^{p-1} = 2^p - 2^{p-1}$$
• Denominator: $$1 + 2^{2p-1} = 1 + (2^p \cdot 2^{p-1})$$
• $$p=1: \tan^{-1}(2^1) - \tan^{-1}(2^0)$$
• $$p=2: \tan^{-1}(2^2) - \tan^{-1}(2^1)$$
• ...
• $$p=11: \tan^{-1}(2^{11}) - \tan^{-1}(2^{10})$$

So, $$T_p = \tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \right) = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$$.

Now, let's expand the sum from $$p=1$$ to $$11$$:

$$\sum_{p=1}^{11} [ \tan^{-1}(2^p) - \tan^{-1}(2^{p-1}) ]$$

Expanding the terms:

Notice that all intermediate terms cancel out (telescope), leaving only:

$$\tan^{-1}(2^{11}) - \tan^{-1}(2^0) = \tan^{-1}(2048) - \tan^{-1}(1)$$

Since $$\tan^{-1}(1) = \frac{\pi}{4}$$, the sum is $$\tan^{-1}(2048) - \frac{\pi}{4}$$.

The full equation is:

$$\frac{\pi}{4} + \left( \tan^{-1}(2048) - \frac{\pi}{4} \right) = \alpha$$

$$\tan^{-1}(2048) = \alpha$$

Taking the tangent of both sides:

$$\tan(\tan^{-1}(2048)) = \tan \alpha$$

$$2048 = \tan \alpha$$

Final Answer: 2048