`S_(1),S_(2), S_(3),...,S_(n)` are sums of n infinite geometric progressions. The first terms of these progressions are `1, 2^(2)-1, 2^(3)-1, ..., 2^(n) – 1` and the common ratios are `(1)/(2),(1)/(2^(2)),(1)/(2^(3)), ...., (1)/(2^(n))`. Calculate the value of `S_(1), +S_(2),+ ... + S_(n).`

Answer

To solve the problem, we need to find the sum \( S_1 + S_2 + S_3 + \ldots + S_n \) where each \( S_k \) is the sum of an infinite geometric progression.

### Step 1: Identify the first term and common ratio for each \( S_k \)

The first terms of the progressions are given as:
- \( a_1 = 1 \)
- \( a_2 = 2^2 - 1 = 3 \)
- \( a_3 = 2^3 - 1 = 7 \)
- ...
- \( a_n = 2^n - 1 \)

The common ratios are:
- \( r_1 = \frac{1}{2} \)
- \( r_2 = \frac{1}{2^2} = \frac{1}{4} \)
- \( r_3 = \frac{1}{2^3} = \frac{1}{8} \)
- ...
- \( r_n = \frac{1}{2^n} \)

### Step 2: Use the formula for the sum of an infinite geometric series

The sum \( S_k \) of an infinite geometric series can be calculated using the formula:
\[ S_k = \frac{a_k}{1 - r_k} \]
where \( a_k \) is the first term and \( r_k \) is the common ratio.

### Step 3: Calculate \( S_k \) for each \( k \)

1. **For \( S_1 \)**:
\[ S_1 = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

2. **For \( S_2 \)**:
\[ S_2 = \frac{2^2 - 1}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4 \]

3. **For \( S_3 \)**:
\[ S_3 = \frac{2^3 - 1}{1 - \frac{1}{8}} = \frac{7}{\frac{7}{8}} = 8 \]

4. **For \( S_4 \)**:
\[ S_4 = \frac{2^4 - 1}{1 - \frac{1}{16}} = \frac{15}{\frac{15}{16}} = 16 \]

Continuing this pattern, we can see that:
\[ S_k = 2^k \quad \text{for } k = 1, 2, 3, \ldots, n \]

### Step 4: Sum \( S_1 + S_2 + S_3 + \ldots + S_n \)

Now we need to calculate:
\[ S_1 + S_2 + S_3 + \ldots + S_n = 2 + 4 + 8 + \ldots + 2^n \]

This is a geometric series where the first term \( a = 2 \) and the common ratio \( r = 2 \). The sum of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Substituting the values:
\[ S_n = 2 \frac{2^n - 1}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2 \]

### Final Answer

Thus, the value of \( S_1 + S_2 + S_3 + \ldots + S_n \) is:
\[ \boxed{2^{n+1} - 2} \]