The value of ` 5 * cot ( Sigma_(k =1)^(5) cot ^(-1) ( k^(2) + k + 1))` is equal to

Options

A. `5/2`
B. 7
C. `-7`
D. `7/2`

Answer

To solve the problem, we need to evaluate the expression \( 5 \cdot \cot \left( \sum_{k=1}^{5} \cot^{-1} (k^2 + k + 1) \right) \).

### Step-by-Step Solution:

1. **Rewrite the Summation**:
We start with the summation:
\[ \sum_{k=1}^{5} \cot^{-1} (k^2 + k + 1) \]
We can use the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \). Thus, we can rewrite the summation as:
\[ \sum_{k=1}^{5} \tan^{-1}\left(\frac{1}{k^2 + k + 1}\right) \]

2. **Use the Formula for the Difference of Inverses**:
We know that:
\[ \tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a-b}{1 + ab}\right) \]
We can express \( \tan^{-1}\left(\frac{1}{k^2 + k + 1}\right) \) in terms of \( k \) and \( k+1 \):
\[ \tan^{-1}(k+1) - \tan^{-1}(k) \]
Therefore, our summation becomes:
\[ \sum_{k=1}^{5} \left( \tan^{-1}(k+1) - \tan^{-1}(k) \right) \]

3. **Evaluate the Telescoping Series**:
The series telescopes:
\[ \tan^{-1}(2) - \tan^{-1}(1) + \tan^{-1}(3) - \tan^{-1}(2) + \tan^{-1}(4) - \tan^{-1}(3) + \tan^{-1}(5) - \tan^{-1}(4) + \tan^{-1}(6) - \tan^{-1}(5) \]
Most terms cancel out, leaving us with:
\[ \tan^{-1}(6) - \tan^{-1}(1) \]

4. **Combine the Inverses**:
We can now express this as:
\[ \tan^{-1}\left(\frac{6 - 1}{1 + 6 \cdot 1}\right) = \tan^{-1}\left(\frac{5}{7}\right) \]

5. **Find the Cotangent**:
Now we need to find:
\[ \cot\left(\tan^{-1}\left(\frac{5}{7}\right)\right) \]
Using the identity \( \cot(\tan^{-1}(x)) = \frac{1}{x} \):
\[ \cot\left(\tan^{-1}\left(\frac{5}{7}\right)\right) = \frac{7}{5} \]

6. **Final Calculation**:
Now we can substitute back into our original expression:
\[ 5 \cdot \cot\left(\sum_{k=1}^{5} \cot^{-1}(k^2 + k + 1)\right) = 5 \cdot \frac{7}{5} = 7 \]

### Conclusion:
Thus, the value of \( 5 \cdot \cot\left(\sum_{k=1}^{5} \cot^{-1}(k^2 + k + 1)\right) \) is \( \boxed{7} \).