If (π)/(4) + ∑_ p=1 ^ 11 ^ -1 ( 2^ p-1 1 + 2^ 2p-1 ) = α , then α...

The general term of the given series is $T_p = \tan^{-1}\left(\dfrac{2^{p-1}}{1 + 2^{2p-1}}\right)$.

This can be rewritten by expressing the numerator and denominator in terms of $2^p$ and $2^{p-1}$:

$T_p = \tan^{-1}\left(\dfrac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}}\right)$

Using the inverse trigonometric identity $\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1}x - \tan^{-1}y$, we get:

$T_p = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$

Summing $T_p$ from $p = 1$ to $11$ gives a telescoping series:

$\displaystyle\sum_{p=1}^{11} T_p = \left(\tan^{-1}(2^1) - \tan^{-1}(2^0)\right) + \left(\tan^{-1}(2^2) - \tan^{-1}(2^1)\right) + \dots + \left(\tan^{-1}(2^{11}) - \tan^{-1}(2^{10})\right)$

Canceling the intermediate terms, we obtain:

$\displaystyle\sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \tan^{-1}(2^0)$

Since $2^0 = 1$ and $\tan^{-1}(1) = \dfrac{\pi}{4}$, the sum becomes:

$\displaystyle\sum_{p=1}^{11} T_p = \tan^{-1}(2^{11}) - \dfrac{\pi}{4}$

Substituting this back into the expression for $\alpha$:

$\alpha = \dfrac{\pi}{4} + \tan^{-1}(2^{11}) - \dfrac{\pi}{4}$

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

Taking the tangent of both sides, we get:

$\tan\alpha = 2^{11} = 2048$

Answer: $2048$