If $$\sum_{r=1}^{50} \tan^{-1} \frac{1}{2r^2} = p$$, then the value of $$\tan p$$ is:

We have to evaluate the expression

$$p=\sum_{r=1}^{50}\tan^{-1}\frac{1}{2r^{2}}$$

and then obtain $$\tan p.$$

First, recall the subtraction formula for the inverse tangent:

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

We want the right-hand side of this identity to be $$\tan^{-1}\dfrac{1}{2r^{2}}.$$ To achieve that, choose

$$x=2r+1,\qquad y=2r-1.$$

Substituting these in the fraction that appears in the formula gives

$$$\frac{x-y}{1+xy}= \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)} =\frac{2}{1+\left((2r)^{2}-1\right)} =\frac{2}{1+4r^{2}-1} =\frac{2}{4r^{2}} =\frac{1}{2r^{2}}.$$$

Hence

$$$\tan^{-1}\frac{1}{2r^{2}} =\tan^{-1}(2r+1)-\tan^{-1}(2r-1).$$$

We now replace every term in the given sum by this difference:

$$$\begin{aligned} p&=\sum_{r=1}^{50}\left[\tan^{-1}(2r+1)-\tan^{-1}(2r-1)\right]\\[4pt] &=\bigl(\tan^{-1}3-\tan^{-1}1\bigr) +\bigl(\tan^{-1}5-\tan^{-1}3\bigr) +\bigl(\tan^{-1}7-\tan^{-1}5\bigr)\\[2pt] &\ \ \,+\;\cdots\; +\bigl(\tan^{-1}101-\tan^{-1}99\bigr). \end{aligned}$$$

Observe the telescoping nature of this series: every intermediate term cancels with its negative counterpart. All that survives is the first negative term and the last positive term. Therefore

$$p=\tan^{-1}101-\tan^{-1}1.$$

To find $$\tan p$$ we again use the tangent-difference formula, this time for ordinary tangents:

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B}.$$

Here $$A=\tan^{-1}101$$ and $$B=\tan^{-1}1,$$ so $$\tan A=101$$ and $$\tan B=1.$$ Substituting, we get

$$$\tan p =\frac{101-1}{1+101\cdot1} =\frac{100}{102} =\frac{50}{51}.$$$

Hence, the correct answer is Option C.