If `S _(1), S _(2) , S _(3)……., S _(2n)` are the sums of infinite geometric series whose first terms are respectively `1,2,3,…..,2n` and common ratio are respectively,
`1/2, 1/3, …….., (1)/(2n +1),` find the value of ,` S_(1) ^(2) + S_(2) ^(2) +…....+ S _(2n -1) ^(2).`

Answer

To solve the problem, we need to find the value of \( S_1^2 + S_2^2 + S_3^2 + \ldots + S_{2n-1}^2 \) where \( S_k \) represents the sum of an infinite geometric series with first term \( k \) and common ratio \( \frac{1}{k+1} \).

### Step-by-Step Solution:

1. **Identify the formula for the sum of an infinite geometric series**:
The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[ S = \frac{a}{1 - r} \]

2. **Calculate \( S_k \) for \( k = 1, 2, \ldots, 2n \)**:
- For \( S_1 \):
\[ S_1 = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]
- For \( S_2 \):
\[ S_2 = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3 \]
- For \( S_3 \):
\[ S_3 = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4 \]
- Continuing this pattern, we can derive:
\[ S_k = \frac{k}{1 - \frac{1}{k+1}} = \frac{k}{\frac{k}{k+1}} = k + 1 \]
- Therefore, for \( k = 1, 2, \ldots, 2n \):
\[ S_k = k + 1 \]

3. **Calculate \( S_1^2 + S_2^2 + \ldots + S_{2n-1}^2 \)**:
- We need to compute:
\[ S_1^2 + S_2^2 + S_3^2 + \ldots + S_{2n-1}^2 = (2)^2 + (3)^2 + (4)^2 + \ldots + (2n)^2 \]
- This can be rewritten as:
\[ S = 2^2 + 3^2 + 4^2 + \ldots + (2n)^2 \]

4. **Use the formula for the sum of squares**:
The sum of squares of the first \( m \) natural numbers is given by:
\[ \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6} \]
- To find \( \sum_{k=1}^{2n} k^2 \):
\[ \sum_{k=1}^{2n} k^2 = \frac{2n(2n+1)(4n+1)}{6} \]
- To find \( \sum_{k=1}^{1} k^2 \):
\[ \sum_{k=1}^{1} k^2 = 1^2 = 1 \]

5. **Combine the results**:
\[ S = \sum_{k=1}^{2n} k^2 - \sum_{k=1}^{1} k^2 = \frac{2n(2n+1)(4n+1)}{6} - 1 \]

### Final Result:
Thus, the value of \( S_1^2 + S_2^2 + \ldots + S_{2n-1}^2 \) is:
\[ S = \frac{2n(2n+1)(4n+1)}{6} - 1 \]