The sum of the series `cot^(-1)((9)/(2))+cot^(-1)((33)/(4))+cot^(-1)((129)/(8))+…….oo` is equal to :

Options

A. `cot^(-1)(2)`
B. `cot^(-1)(3)`
C. `cot^(-1)(-1)`
D. `cot^(-1)(1)`

Answer

To find the sum of the series \( S = \cot^{-1}\left(\frac{9}{2}\right) + \cot^{-1}\left(\frac{33}{4}\right) + \cot^{-1}\left(\frac{129}{8}\right) + \ldots \), we will use the identity for the difference of inverse cotangents.

### Step 1: Identify the pattern in the series
We observe that the terms in the series can be expressed in terms of the cotangent inverse difference identity:
\[ \cot^{-1}(x) - \cot^{-1}(y) = \cot^{-1}\left(\frac{1 + xy}{y - x}\right) \]
We will rewrite the series in a form that allows us to apply this identity.

### Step 2: Rewrite the terms
The first term is \( \cot^{-1}\left(\frac{9}{2}\right) \). We can express the subsequent terms in a similar manner:
- The second term \( \cot^{-1}\left(\frac{33}{4}\right) \) can be rewritten using the identity.
- The third term \( \cot^{-1}\left(\frac{129}{8}\right) \) can also be rewritten.

### Step 3: Apply the identity
Using the identity, we can express the series as:
\[ S = \cot^{-1}(2) - \cot^{-1}(4) + \cot^{-1}(4) - \cot^{-1}(8) + \cot^{-1}(8) - \cot^{-1}(16) + \ldots \]
This shows that each term cancels with the next, leading to a telescoping series.

### Step 4: Simplify the series
After cancellation, we are left with:
\[ S = \cot^{-1}(2) - \lim_{n \to \infty} \cot^{-1}(2^n) \]
As \( n \to \infty \), \( \cot^{-1}(2^n) \to 0 \).

### Step 5: Final result
Thus, we have:
\[ S = \cot^{-1}(2) \]

### Conclusion
The sum of the series is:
\[ \boxed{\cot^{-1}(2)} \]