If `S_(1), S_(2),cdots S_(n)`, ., are the sums of infinite geometric series whose first terms are 1, 2, 3,…… ,n and common ratios are `(1)/(2) ,(1)/(3),(1)/(4),cdots,(1)/(n+1)` then `S_(1)+S_(2)+S_(3)+cdots+S_(n)`=

Options

A. `(n(n+1))/(2)`
B. `((n+1)(n+3))/(2)`
C. `(n(n+2))/(2)`
D. `(n(n+3))/(2)`

Answer

To solve the problem, we need to find the sum \( S_1 + S_2 + S_3 + \cdots + S_n \), where each \( S_k \) is the sum of an infinite geometric series with the first term \( k \) and a 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 each \( k \)**:
For our series, the first term \( a = k \) and the common ratio \( r = \frac{1}{k+1} \).
Therefore, we can write:
\[ S_k = \frac{k}{1 - \frac{1}{k+1}} = \frac{k}{\frac{k+1 - 1}{k+1}} = \frac{k}{\frac{k}{k+1}} = k \cdot \frac{k+1}{k} = k + 1 \]

3. **Sum \( S_1 + S_2 + S_3 + \cdots + S_n \)**:
Now we need to find:
\[ S_1 + S_2 + S_3 + \cdots + S_n = (1 + 1) + (2 + 1) + (3 + 1) + \cdots + (n + 1) \]
This can be simplified to:
\[ = 2 + 3 + 4 + \cdots + (n + 1) \]

4. **Calculate the sum of the first \( n \) natural numbers**:
The sum of the first \( n \) natural numbers is given by:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
Thus, the sum from \( 2 \) to \( n + 1 \) can be calculated as:
\[ \text{Sum from } 1 \text{ to } (n + 1) - 1 = \frac{(n + 1)(n + 2)}{2} - 1 \]

5. **Final Calculation**:
The total sum becomes:
\[ S_1 + S_2 + S_3 + \cdots + S_n = \frac{(n + 1)(n + 2)}{2} - 1 \]
Simplifying this gives:
\[ = \frac{n(n + 1)}{2} + (n + 1) - 1 = \frac{n(n + 1)}{2} + n = \frac{n(n + 1) + 2n}{2} = \frac{n(n + 1 + 2)}{2} = \frac{n(n + 3)}{2} \]

Thus, the final answer is:
\[ S_1 + S_2 + S_3 + \cdots + S_n = \frac{n(n + 3)}{2} \]