$$S = \tan^{-1}\left(\frac{1}{n^2+n+1}\right) + \tan^{-1}\left(\frac{1}{n^2+3n+3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$$, then $$\tan S$$ is equal to :

The given sum is:

$$S = \tan^{-1}\left(\frac{1}{n^2+n+1}\right) + \tan^{-1}\left(\frac{1}{n^2+3n+3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$$

First, we identify the general term. The last term has the denominator $$1 + (n+19)(n+20) = 1 + n^2 + 39n + 380 = n^2 + 39n + 381$$. Observing the denominators:

• The first term has denominator $$n^2 + n + 1$$.
• The second term has denominator $$n^2 + 3n + 3$$.

We notice that for each term, the denominator is of the form $$n^2 + (2k+1)n + (k^2 + k + 1)$$ where $$k$$ ranges from 0 to 19. Let's verify:

• For $$k = 0$$: $$n^2 + (2 \cdot 0 + 1)n + (0^2 + 0 + 1) = n^2 + n + 1$$, which matches the first term.
• For $$k = 1$$: $$n^2 + (2 \cdot 1 + 1)n + (1^2 + 1 + 1) = n^2 + 3n + 3$$, which matches the second term.
• For $$k = 19$$: $$n^2 + (2 \cdot 19 + 1)n + (19^2 + 19 + 1) = n^2 + 39n + (361 + 19 + 1) = n^2 + 39n + 381$$, which matches the last term.

Thus, the sum has 20 terms, from $$k = 0$$ to $$k = 19$$:

$$S = \sum_{k=0}^{19} \tan^{-1}\left(\frac{1}{n^2 + (2k+1)n + (k^2 + k + 1)}\right)$$

We can rewrite the denominator as:

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

So each term becomes:

$$\tan^{-1}\left(\frac{1}{(n + k)^2 + (n + k) + 1}\right)$$

Now, recall the identity:

$$\tan^{-1}\left(\frac{1}{m^2 + m + 1}\right) = \tan^{-1}(m + 1) - \tan^{-1}(m)$$

This is verified by setting $$\theta = \tan^{-1}(m + 1) - \tan^{-1}(m)$$, so:

$$\tan \theta = \frac{(m + 1) - m}{1 + (m + 1)m} = \frac{1}{1 + m(m + 1)} = \frac{1}{m^2 + m + 1}$$

Thus, $$\theta = \tan^{-1}\left(\frac{1}{m^2 + m + 1}\right)$$.

Applying this identity with $$m = n + k$$, each term simplifies to:

$$\tan^{-1}\left(\frac{1}{(n + k)^2 + (n + k) + 1}\right) = \tan^{-1}((n + k) + 1) - \tan^{-1}(n + k) = \tan^{-1}(n + k + 1) - \tan^{-1}(n + k)$$

Therefore, the sum $$S$$ becomes:

$$S = \sum_{k=0}^{19} \left[ \tan^{-1}(n + k + 1) - \tan^{-1}(n + k) \right]$$

This is a telescoping series. Writing out the terms:

• For $$k = 0$$: $$\tan^{-1}(n + 1) - \tan^{-1}(n)$$
• For $$k = 1$$: $$\tan^{-1}(n + 2) - \tan^{-1}(n + 1)$$
• For $$k = 2$$: $$\tan^{-1}(n + 3) - \tan^{-1}(n + 2)$$
• $$\vdots$$
• For $$k = 19$$: $$\tan^{-1}(n + 20) - \tan^{-1}(n + 19)$$

Summing these, intermediate terms cancel:

$$S = -\tan^{-1}(n) + \tan^{-1}(n + 20)$$

Because $$\tan^{-1}(n + 1)$$ cancels with $$-\tan^{-1}(n + 1)$$, $$\tan^{-1}(n + 2)$$ cancels with $$-\tan^{-1}(n + 2)$$, and so on, up to $$\tan^{-1}(n + 19)$$.

So:

$$S = \tan^{-1}(n + 20) - \tan^{-1}(n)$$

Now, we need to find $$\tan S = \tan\left( \tan^{-1}(n + 20) - \tan^{-1}(n) \right)$$. Using the tangent subtraction formula:

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

Here, $$A = \tan^{-1}(n + 20)$$, so $$\tan A = n + 20$$, and $$B = \tan^{-1}(n)$$, so $$\tan B = n$$. Substituting:

$$\tan S = \frac{(n + 20) - n}{1 + (n + 20) \cdot n} = \frac{20}{1 + n(n + 20)}$$

Simplifying the denominator:

$$1 + n(n + 20) = 1 + n^2 + 20n = n^2 + 20n + 1$$

Thus:

$$\tan S = \frac{20}{n^2 + 20n + 1}$$

Comparing with the options:

• A. $$\frac{20}{401 + 20n}$$
• B. $$\frac{n}{n^2 + 20n + 1}$$
• C. $$\frac{20}{n^2 + 20n + 1}$$
• D. $$\frac{n}{401 + 20n}$$

Option C matches the expression we obtained.

Hence, the correct answer is Option C.