Choose the proposition among the following that is true for all `n in NN`

Options

A. (a) `tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+...+tan^(_1)((1)/(n^(2)+n+1))=tan^(-1)((3n-1)/(3n+3))`
B. (b) `tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+"tan"^(-1)((1)/(n^(2)+n+1))=tan^(-1)((3n-1)/(4n+2))`
C. (c) `tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+"tan"^(-1)((1)/(n^(2)+n+1))=tan^(-1)((n)/(n+2))`
D. (d) `tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+"tan"^(-1)((1)/(n^(2)+n+1))=tan^(-1)((2n-1)/(3n))`

Answer

To solve the problem, we need to evaluate the sum of the series given in the question and compare it with the provided options to determine which proposition is true for all natural numbers \( n \).

### Step-by-Step Solution

1. **Understanding the Series**: We need to evaluate the sum:
\[ S_n = \sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{r^2 + r + 1}\right) \]

2. **Simplifying the General Term**: The general term can be rewritten as:
\[ \tan^{-1}\left(\frac{1}{r^2 + r + 1}\right) = \tan^{-1}\left(\frac{1}{1 + r(r + 1)}\right) \]
We can express \( 1 \) in the numerator as \( (r + 1) - r \):
\[ = \tan^{-1}\left(\frac{(r + 1) - r}{1 + r(r + 1)}\right) \]

3. **Using the Tan Inverse Identity**: We apply the identity:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
This gives us:
\[ \tan^{-1}(r + 1) - \tan^{-1}(r) \]

4. **Rewriting the Sum**: Thus, we can rewrite the sum \( S_n \):
\[ S_n = \sum_{r=1}^{n} \left(\tan^{-1}(r + 1) - \tan^{-1}(r)\right) \]

5. **Telescoping Series**: This is a telescoping series, where most terms cancel out:
\[ S_n = \tan^{-1}(n + 1) - \tan^{-1}(1) \]

6. **Calculating the Result**: Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have:
\[ S_n = \tan^{-1}(n + 1) - \frac{\pi}{4} \]

7. **Final Expression**: We can express this as:
\[ S_n = \tan^{-1}\left(\frac{n + 1 - 1}{1 + (n + 1) \cdot 1}\right) = \tan^{-1}\left(\frac{n}{n + 2}\right) \]

8. **Comparing with Options**: Now we compare \( S_n \) with the given options:
- Option A: \( \tan^{-1}\left(\frac{3n - 1}{3n + 3}\right) \)
- Option B: \( \tan^{-1}\left(\frac{3n - 1}{4n + 2}\right) \)
- Option C: \( \tan^{-1}\left(\frac{n}{n + 2}\right) \)
- Option D: \( \tan^{-1}\left(\frac{2n - 1}{3n}\right) \)

We find that:
\[ S_n = \tan^{-1}\left(\frac{n}{n + 2}\right) \]
matches with **Option C**.

### Conclusion
Thus, the correct answer is **Option C: \( \tan^{-1}\left(\frac{n}{n + 2}\right) \)**.