The value of $$\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$$ is

We need to find $$\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$$. Note that $$1 + n + n^2 = 1 + n(1 + n)$$, so $$\frac{1}{1 + n + n^2} = \frac{(n+1) - n}{1 + n(n+1)}$$.

Using the inverse tangent subtraction formula $$\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}$$, we get $$\tan^{-1}\frac{(n+1) - n}{1 + n(n+1)} = \tan^{-1}(n+1) - \tan^{-1}(n)$$. Hence the sum telescopes:

$$\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right) = \sum_{n=1}^{50} [\tan^{-1}(n+1) - \tan^{-1}(n)] = \tan^{-1}(51) - \tan^{-1}(1) = \tan^{-1}(51) - \frac{\pi}{4}$$.

Let $$\theta = \tan^{-1}(51) - \frac{\pi}{4}$$. Then $$\tan\theta = \tan\left(\tan^{-1}(51) - \frac{\pi}{4}\right) = \frac{51 - 1}{1 + 51 \cdot 1} = \frac{50}{52} = \frac{25}{26}$$, so $$\cot\theta = \frac{1}{\tan\theta} = \frac{26}{25}$$.

The correct answer is Option C: $$\frac{26}{25}$$.