Let `A_(1),A_(2),A_(3),…, A_(100)` be 100 seta and such that `n(A_(1))=i+1and A_(1)subA_(2)subA_(3)sub...A_(100),then underset(i=1)overset(100)UAi` contains… elements

Options

A. 99
B. 100
C. 101
D. 102

Answer

To solve the problem, we need to find the number of elements in the union of the sets \( A_1, A_2, A_3, \ldots, A_{100} \) given the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Sets**:
We are given that \( n(A_i) = i + 1 \) for \( i = 1, 2, \ldots, 100 \). This means:
- \( n(A_1) = 1 + 1 = 2 \)
- \( n(A_2) = 2 + 1 = 3 \)
- \( n(A_3) = 3 + 1 = 4 \)
- ...
- \( n(A_{100}) = 100 + 1 = 101 \)

2. **Subset Relationships**:
The problem states that \( A_1 \subset A_2 \subset A_3 \subset \ldots \subset A_{100} \). This means that every element in \( A_1 \) is also in \( A_2 \), every element in \( A_2 \) is in \( A_3 \), and so on, up to \( A_{100} \). Therefore, \( A_{100} \) contains all the elements from all the previous sets.

3. **Finding the Union**:
The union of all the sets can be expressed as:
\[ A_1 \cup A_2 \cup A_3 \cup \ldots \cup A_{100} = A_{100} \]
This is because \( A_{100} \) contains all the elements from \( A_1 \) to \( A_{99} \).

4. **Calculating the Number of Elements in \( A_{100} \)**:
Since we know that \( n(A_{100}) = 100 + 1 = 101 \), we can conclude:
\[ n(A_1 \cup A_2 \cup A_3 \cup \ldots \cup A_{100}) = n(A_{100}) = 101 \]

5. **Final Answer**:
Therefore, the number of elements in the union \( \bigcup_{i=1}^{100} A_i \) is \( 101 \).

### Conclusion:
The final answer is:
\[ \text{The union } \bigcup_{i=1}^{100} A_i \text{ contains } 101 \text{ elements.} \]