The sequence of sets `S_(1), S_(2),S_(3),S_(4),…..` is defined as `S_(1) = {1}. S_(2) = {3,5}, S_(3) = {7,9,11}, S_(4) = {13,15,17,19}`….etc.
The last element of the set `S_(100)` is:

Options

A. 10099
B. 9899
C. 9900
D. none of these

Answer

To find the last element of the set \( S_{100} \), we can follow these steps:

### Step 1: Identify the pattern in the sets
The sets are defined as follows:
- \( S_1 = \{1\} \) (1 element)
- \( S_2 = \{3, 5\} \) (2 elements)
- \( S_3 = \{7, 9, 11\} \) (3 elements)
- \( S_4 = \{13, 15, 17, 19\} \) (4 elements)

From this, we can observe that:
- The number of elements in \( S_n \) is \( n \).
- All elements in these sets are odd numbers.

### Step 2: Find the first element of each set
We can find the first element of each set using a formula derived from the pattern:
- The first element of \( S_n \) can be expressed as \( 2n^2 - 2n + 1 \).

Let's verify this for the first few sets:
- For \( n = 1 \): \( 2(1^2) - 2(1) + 1 = 1 \) (correct)
- For \( n = 2 \): \( 2(2^2) - 2(2) + 1 = 3 \) (correct)
- For \( n = 3 \): \( 2(3^2) - 2(3) + 1 = 7 \) (correct)
- For \( n = 4 \): \( 2(4^2) - 2(4) + 1 = 13 \) (correct)

### Step 3: Calculate the first element of \( S_{100} \)
Using the formula for \( n = 100 \):
\[ \text{First element of } S_{100} = 2(100^2) - 2(100) + 1 = 20000 - 200 + 1 = 19801 \]

### Step 4: Find the last element of \( S_{100} \)
Since \( S_{100} \) contains 100 elements and the elements are in an arithmetic progression (AP) with a common difference of 2, we can find the last element using the formula for the \( n \)-th term of an AP:
\[ \text{Last element} = \text{First element} + (n - 1) \cdot d \]
Where:
- First element = 19801
- \( n = 100 \)
- \( d = 2 \)

Calculating the last element:
\[ \text{Last element} = 19801 + (100 - 1) \cdot 2 = 19801 + 99 \cdot 2 = 19801 + 198 = 20000 \]

### Conclusion
The last element of the set \( S_{100} \) is \( \boxed{20000} \).