If `S_(1), S_(2),…S_(lambda)` are the sums of infinite G.P.'s whose first terms are respectively 1, 2, 3,….`lambda` and common ratios are `(1)/(2), (1)/(3),…(1)/(lambda + 1)` respectively, then `S_(1) + S_(2) + S_(3) +...+ S_(lambda) =`

Options

A. `(lambda (lambda + 1))/(2)`
B. `(lambda (lambda + 2))/(2)`
C. `(lambda (lambda + 3))/(2)`
D. none

Answer

To solve the problem, we need to find the sum \( S_1 + S_2 + S_3 + \ldots + S_\lambda \) where each \( S_i \) is the sum of an infinite geometric progression (G.P.) with specific first terms and common ratios.

### Step-by-Step Solution:

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

2. **Calculate \( S_1 \)**:
For \( S_1 \), the first term \( a = 1 \) and the common ratio \( r = \frac{1}{2} \):
\[ S_1 = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 \]

3. **Calculate \( S_2 \)**:
For \( S_2 \), the first term \( a = 2 \) and the common ratio \( r = \frac{1}{3} \):
\[ S_2 = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3 \]

4. **Calculate \( S_3 \)**:
For \( S_3 \), the first term \( a = 3 \) and the common ratio \( r = \frac{1}{4} \):
\[ S_3 = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4 \]

5. **Calculate \( S_4 \)**:
For \( S_4 \), the first term \( a = 4 \) and the common ratio \( r = \frac{1}{5} \):
\[ S_4 = \frac{4}{1 - \frac{1}{5}} = \frac{4}{\frac{4}{5}} = 5 \]

6. **Generalize for \( S_\lambda \)**:
Continuing this pattern, we can see that:
\[ S_i = \frac{i}{1 - \frac{1}{i + 1}} = \frac{i}{\frac{i}{i + 1}} = i + 1 \]
Therefore, we can generalize:
\[ S_\lambda = \lambda + 1 \]

7. **Sum \( S_1 + S_2 + S_3 + \ldots + S_\lambda \)**:
Now, we sum all the values:
\[ S_1 + S_2 + S_3 + \ldots + S_\lambda = 2 + 3 + 4 + \ldots + (\lambda + 1) \]
This is an arithmetic series where:
- First term \( a = 2 \)
- Last term \( l = \lambda + 1 \)
- Number of terms \( n = \lambda \)

The sum of an arithmetic series is given by:
\[ S_n = \frac{n}{2} \times (a + l) \]
Thus:
\[ S = \frac{\lambda}{2} \times (2 + (\lambda + 1)) = \frac{\lambda}{2} \times (\lambda + 3) \]

### Final Result:
The final sum \( S_1 + S_2 + S_3 + \ldots + S_\lambda \) is:
\[ \frac{\lambda(\lambda + 3)}{2} \]