The value of $\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right)$ is

The value of cot(∑ₙ₌₁²³cot⁻¹(1+∑ₖ₌₁ⁿ2k)) is

$\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{k=1}^{n}2k\right)\right)$

$\text{To find: the value of this expression}$

$\sum_{k=1}^{n} 2k = 2\cdot\dfrac{n(n+1)}{2} = n(n+1)$

$1 + \sum_{k=1}^{n} 2k = n^2 + n + 1$

$\cot^{-1}(n^2+n+1) = \tan^{-1}\!\left(\dfrac{1}{n^2+n+1}\right)$

$n^2+n+1 = 1 + n(n+1)$

$\tan^{-1}\dfrac{(n+1)-n}{1+n(n+1)} = \tan^{-1}(n+1) - \tan^{-1}(n)$

$\sum_{n=1}^{23}\cot^{-1}(n^2+n+1) = \Big[\tan^{-1}(n+1)-\tan^{-1}(n)\Big]_{n=1}^{23}$

$= \tan^{-1}24 - \tan^{-1}1 = \tan^{-1}24 - \dfrac{\pi}{4}$

$\theta = \tan^{-1}24 - \dfrac{\pi}{4}$

$\tan\theta = \dfrac{24-1}{1+24\cdot 1} = \dfrac{23}{25}$

$\cot\theta = \dfrac{25}{23}$

$\boxed{\text{Answer: option (2)}\ \ \dfrac{25}{23}}$