If `a_(n) = sqrt(1+(1+(1)/(n))^(2))+sqrt(1+(1-(1)/(n))^(2))` then value of `sum_(n=1)^(20) (1)/(a^(n))` is

Options

A. even
B. odd
C. prime
D. divisible by `'3'`

Explanation

To solve the problem, we need to evaluate the expression for \( a_n \) and then compute the summation \( \sum_{n=1}^{20} \frac{1}{a_n} \).

### Step-by-Step Solution:

1. **Define \( a_n \)**:
\[ a_n = \sqrt{1 + \left(1 + \frac{1}{n}\right)^2} + \sqrt{1 + \left(1 - \frac{1}{n}\right)^2} \]

2. **Expand the squares inside the square roots**:
\[ \left(1 + \frac{1}{n}\right)^2 = 1 + 2 \cdot \frac{1}{n} + \left(\frac{1}{n}\right)^2 = 1 + \frac{2}{n} + \frac{1}{n^2} \]
\[ \left(1 - \frac{1}{n}\right)^2 = 1 - 2 \cdot \frac{1}{n} + \left(\frac{1}{n}\right)^2 = 1 - \frac{2}{n} + \frac{1}{n^2} \]

3. **Substituting back into \( a_n \)**:
\[ a_n = \sqrt{1 + 1 + \frac{2}{n} + \frac{1}{n^2}} + \sqrt{1 + 1 - \frac{2}{n} + \frac{1}{n^2}} \]
\[ = \sqrt{2 + \frac{2}{n} + \frac{1}{n^2}} + \sqrt{2 - \frac{2}{n} + \frac{1}{n^2}} \]

4. **Simplify each square root**:
\[ = \sqrt{2 + \frac{2}{n} + \frac{1}{n^2}} + \sqrt{2 - \frac{2}{n} + \frac{1}{n^2}} \]

5. **Combine the terms**:
To simplify further, we can factor out common terms:
\[ = \sqrt{2 + \frac{1}{n^2}} + \sqrt{2 + \frac{1}{n^2}} = 2\sqrt{2 + \frac{1}{n^2}} \]

6. **Evaluate \( \frac{1}{a_n} \)**:
\[ \frac{1}{a_n} = \frac{1}{2\sqrt{2 + \frac{1}{n^2}}} \]

7. **Set up the summation**:
\[ \sum_{n=1}^{20} \frac{1}{a_n} = \sum_{n=1}^{20} \frac{1}{2\sqrt{2 + \frac{1}{n^2}}} \]
\[ = \frac{1}{2} \sum_{n=1}^{20} \frac{1}{\sqrt{2 + \frac{1}{n^2}}} \]

8. **Approximate the summation**:
For large \( n \), \( \sqrt{2 + \frac{1}{n^2}} \) approaches \( \sqrt{2} \). Thus:
\[ \sum_{n=1}^{20} \frac{1}{\sqrt{2 + \frac{1}{n^2}}} \approx \sum_{n=1}^{20} \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \]

9. **Final calculation**:
\[ \sum_{n=1}^{20} \frac{1}{a_n} \approx \frac{1}{2} \cdot 10\sqrt{2} = 5\sqrt{2} \]

### Final Answer:
The value of \( \sum_{n=1}^{20} \frac{1}{a_n} \) is approximately \( 5\sqrt{2} \).