If $$\cot^{-1}(\alpha) = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + \ldots$$ upto 100 terms, then $$\alpha$$ is:

We need to find $$\alpha$$ where $$\cot^{-1}(\alpha) = \cot^{-1}2 + \cot^{-1}8 + \cot^{-1}18 + \cot^{-1}32 + \ldots$$ up to 100 terms.

The general term has the pattern: $$2, 8, 18, 32, \ldots$$. The $$n$$-th term is $$2n^2$$ (since $$2(1)^2 = 2$$, $$2(2)^2 = 8$$, $$2(3)^2 = 18$$, $$2(4)^2 = 32$$).

We use the identity: $$\cot^{-1}(2n^2) = \tan^{-1}\frac{1}{2n^2}$$. We can write $$\frac{1}{2n^2} = \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$$, since the denominator gives $$1 + 4n^2 - 1 = 4n^2$$ — let us verify: $$\frac{2}{4n^2} = \frac{1}{2n^2}$$. Yes, this works.

By the identity $$\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}$$, we get $$\tan^{-1}\frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$$.

So $$\cot^{-1}(2n^2) = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$$.

Summing from $$n = 1$$ to $$100$$, this is a telescoping series: $$\sum_{n=1}^{100} [\tan^{-1}(2n+1) - \tan^{-1}(2n-1)] = \tan^{-1}(201) - \tan^{-1}(1)$$.

So $$\cot^{-1}(\alpha) = \tan^{-1}(201) - \tan^{-1}(1) = \tan^{-1}\frac{201 - 1}{1 + 201 \cdot 1} = \tan^{-1}\frac{200}{202} = \tan^{-1}\frac{100}{101}$$.

Since $$\cot^{-1}(\alpha) = \tan^{-1}\frac{100}{101}$$, and $$\cot^{-1}(\alpha) = \tan^{-1}\frac{1}{\alpha}$$, we get $$\frac{1}{\alpha} = \frac{100}{101}$$, so $$\alpha = \frac{101}{100} = 1.01$$.

This matches Option A: $$1.01$$.