`Cot^-1(alpha)=Cot^-1(2)+Cot^-1(8)+Cot^-1(18)+Cot^-1(32)`+ . . then `alpha`=?

Options

A. 1.03
B. 1
C. 1.01
D. 1.02

Answer

To solve the equation \( \cot^{-1}(\alpha) = \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \cot^{-1}(32) + \ldots \), we will use the properties of inverse cotangent and the series given.

### Step-by-Step Solution:

1. **Understanding the Series**:
We are given a series of the form \( \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \cot^{-1}(32) + \ldots \). We need to identify a pattern in the terms.

2. **Identifying the Pattern**:
The terms can be rewritten as:
- \( \cot^{-1}(2) = \cot^{-1}\left(\frac{2}{1^2}\right) \)
- \( \cot^{-1}(8) = \cot^{-1}\left(\frac{2 \cdot 2^2}{2^2}\right) \)
- \( \cot^{-1}(18) = \cot^{-1}\left(\frac{2 \cdot 3^2}{3^2}\right) \)
- \( \cot^{-1}(32) = \cot^{-1}\left(\frac{2 \cdot 4^2}{4^2}\right) \)

This suggests a general term of the form \( \cot^{-1}(2n^2) \) for \( n = 1, 2, 3, \ldots \).

3. **Using the Cotangent Addition Formula**:
We can use the identity:
\[ \cot^{-1}(x) + \cot^{-1}(y) = \cot^{-1}\left(\frac{xy - 1}{x + y}\right) \]
to combine the terms.

4. **Rewriting the Series**:
We can express the series as:
\[ \cot^{-1}(2) + \cot^{-1}(8) + \cot^{-1}(18) + \cot^{-1}(32) + \ldots = \sum_{n=1}^{\infty} \cot^{-1}(2n^2) \]

5. **Converting to Tangent**:
We can convert the cotangent terms to tangent:
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
Thus, we rewrite the series:
\[ \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{2n^2}\right) \]

6. **Finding the Sum**:
The sum can be evaluated using the formula for the sum of arctangents:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
However, for an infinite series, we can recognize that the series converges to a specific value.

7. **Evaluating the Infinite Series**:
The series converges to \( \frac{\pi}{4} \) based on the behavior of the terms as \( n \) approaches infinity.

8. **Setting the Equation**:
We have:
\[ \cot^{-1}(\alpha) = \frac{\pi}{4} \]

9. **Finding \( \alpha \)**:
From the cotangent function, we know:
\[ \cot\left(\frac{\pi}{4}\right) = 1 \]
Therefore, we conclude:
\[ \alpha = 1 \]

### Final Answer:
\[ \alpha = 1 \]