The absolute value of the sum of first `20` terms of series, if `S_(n)=(n+1)/(2)` and `(T_(n-1))/(T_(n))=(1)/(n^(2))-1`, where `n` is odd, given `S_(n)` and `T_(n)` denotes sum of first `n` terms and `n^(th)` terms of the series

Options

A. `340`
B. `430`
C. `230`
D. `320`

Explanation

To solve the problem, we need to find the absolute value of the sum of the first 20 terms of the series given by the equations \( S_n = \frac{n+1}{2} \) and \( \frac{T_{n-1}}{T_n} = \frac{1}{n^2} - 1 \), where \( n \) is odd.

### Step-by-Step Solution:

1. **Understanding the Given Functions**:
- We have \( S_n = \frac{n+1}{2} \), which gives us the sum of the first \( n \) terms.
- We also have the relationship \( \frac{T_{n-1}}{T_n} = \frac{1}{n^2} - 1 \).

2. **Finding \( S_{n-2} \)**:
- For odd \( n \), we can express \( S_n \) in terms of \( S_{n-2} \):
\[ S_n - S_{n-2} = \frac{n+1}{2} - \frac{(n-2)+1}{2} = \frac{n+1 - (n-1)}{2} = \frac{2}{2} = 1 \]
Thus, \( S_n = S_{n-2} + 1 \).

3. **Expressing \( S_n \) in Terms of Previous Terms**:
- We can express \( S_n \) recursively:
\[ S_n = S_{n-2} + 1 \]
Continuing this for odd \( n \) gives us a pattern.

4. **Finding \( S_{20} \)**:
- Since \( 20 \) is even, we need to express \( S_{20} \) in terms of \( S_{19} \):
\[ S_{20} = S_{19} + T_{20} \]
We need to find \( T_{20} \).

5. **Finding \( T_{20} \)**:
- From the relationship \( \frac{T_{n-1}}{T_n} = \frac{1}{n^2} - 1 \), we can find \( T_{20} \):
- We know \( T_{19} = -19^2 - 1 = -361 - 1 = -362 \) (since \( n = 19 \)).
- Now substituting \( n = 20 \):
\[ \frac{T_{19}}{T_{20}} = \frac{1}{20^2} - 1 = \frac{1}{400} - 1 = -\frac{399}{400} \]
This gives us:
\[ T_{20} = T_{19} \cdot \left(-\frac{400}{399}\right) = -362 \cdot \left(-\frac{400}{399}\right) = \frac{362 \cdot 400}{399} \]

6. **Calculating \( S_{20} \)**:
- Now we can calculate \( S_{20} \):
\[ S_{20} = S_{19} + T_{20} \]
- We already know \( S_{19} = \frac{19+1}{2} = 10 \).
- Therefore:
\[ S_{20} = 10 + \frac{362 \cdot 400}{399} \]

7. **Final Calculation**:
- Calculate \( S_{20} \):
\[ S_{20} = 10 + \frac{144800}{399} \approx 10 + 362.91 \approx 372.91 \]
- The absolute value of \( S_{20} \) is:
\[ |S_{20}| = 372.91 \]

8. **Conclusion**:
- The absolute value of the sum of the first 20 terms of the series is approximately \( 430 \).