Evaluate Finite Sum of Inverse Cotangent Series

To find the value of the expression $E = \cot \left( \sum_{n = 1}^{23} \cot^{-1} \left( 1 + \sum_{k = 1}^n 2k \right) \right)$

First, simplify the inner summation: $\sum_{k = 1}^n 2k = 2 \sum_{k = 1}^n k = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$

Substitute this back into the general term $T_n$ of the series: $T_n = \cot^{-1} \left( 1 + n(n+1) \right)$

Using the identity $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$ for $x > 0$, we get: $T_n = \tan^{-1} \left( \frac{1}{1 + n(n+1)} \right)$

Rewrite the argument using the identity $\tan^{-1}\left(\frac{x - y}{1 + xy}\right) = \tan^{-1}(x) - \tan^{-1}(y)$: $T_n = \tan^{-1} \left( \frac{(n+1) - n}{1 + n(n+1)} \right) = \tan^{-1}(n+1) - \tan^{-1}(n)$

Now, sum $T_n$ from $n = 1$ to $23$: $S = \sum_{n = 1}^{23} T_n = \sum_{n = 1}^{23} \Big( \tan^{-1}(n+1) - \tan^{-1}(n) \Big)$

This is a telescoping sum: $S = \big( \tan^{-1}(2) - \tan^{-1}(1) \big) + \big( \tan^{-1}(3) - \tan^{-1}(2) \big) + \dots + \big( \tan^{-1}(24) - \tan^{-1}(23) \big)$ $S = \tan^{-1}(24) - \tan^{-1}(1)$

Using the subtraction formula for tangent, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$: $\tan(S) = \tan\big(\tan^{-1}(24) - \tan^{-1}(1)\big) = \frac{24 - 1}{1 + 24 \cdot 1} = \frac{23}{25}$

Finally, compute the value of the given expression: $E = \cot(S) = \frac{1}{\tan(S)} = \frac{25}{23}$

Hence, the correct option is (B).