If `sum_(n=0)^oo2cot^(-1) ((n^2+n-4)/2)=kpi` then find the value of k

Answer

To solve the problem, we need to evaluate the infinite sum:

\[ \sum_{n=0}^{\infty} 2 \cot^{-1} \left( \frac{n^2 + n - 4}{2} \right) = k\pi \]

### Step 1: Rewrite the cotangent inverse
We know that:

\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]

Thus, we can rewrite the term inside the sum:

\[ \cot^{-1}\left(\frac{n^2 + n - 4}{2}\right) = \tan^{-1}\left(\frac{2}{n^2 + n - 4}\right) \]

### Step 2: Simplify the argument
Next, we simplify the expression:

\[ \frac{2}{n^2 + n - 4} \]

This can be factored or rewritten in a more manageable form. The quadratic \(n^2 + n - 4\) can be expressed as:

\[ n^2 + n - 4 = (n - 2)(n + 2) + 2 \]

However, for our purposes, we can directly work with the original form.

### Step 3: Use the identity for the sum
We can use the identity for the difference of inverse tangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]

By manipulating the terms, we can express the series in terms of a telescoping series.

### Step 4: Telescoping series
The sum can be expressed as:

\[ \sum_{n=0}^{\infty} \left( \tan^{-1}\left(\frac{1}{n + 2}\right) - \tan^{-1}\left(\frac{1}{n + 1}\right) \right) \]

This will telescope, meaning that most terms will cancel out.

### Step 5: Evaluate the limit
As \(n\) approaches infinity, we find:

\[ \lim_{n \to \infty} \tan^{-1}\left(\frac{1}{n + 2}\right) = 0 \]

The first term of the series when \(n=0\) gives us:

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

Thus, the entire sum converges to:

\[ \frac{\pi}{4} \]

### Step 6: Multiply by 2
Since we have a factor of 2 in front of the cotangent inverse, we multiply the result by 2:

\[ 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

### Step 7: Relate to \(k\pi\)
Now we have:

\[ \frac{\pi}{2} = k\pi \]

From this, we can solve for \(k\):

\[ k = \frac{1}{2} \]

### Final Answer
Thus, the value of \(k\) is:

\[ \boxed{\frac{1}{2}} \]