The numbers `1,3,6,10,15,21,28"..."` are called triangular numbers. Let `t_(n)` denote the `n^(th)` triangular number such that `t_(n)=t_(n-1)+n,AA n ge 2`.
The value of `t_(50)` is:

Options

A. (a) `1075`
B. (b) `1175`
C. (c) `1275`
D. (d) `1375`

Explanation

To find the value of the 50th triangular number \( t_{50} \), we will follow the recursive definition of triangular numbers and derive a formula for \( t_n \).

### Step-by-Step Solution:

1. **Understanding the Recursive Definition**:
The triangular numbers are defined recursively as:
\[ t_n = t_{n-1} + n \quad \text{for } n \geq 2 \]
with the base case:
\[ t_1 = 1 \]

2. **Finding the First Few Triangular Numbers**:
Let's compute the first few triangular numbers to understand the pattern:
- \( t_1 = 1 \)
- \( t_2 = t_1 + 2 = 1 + 2 = 3 \)
- \( t_3 = t_2 + 3 = 3 + 3 = 6 \)
- \( t_4 = t_3 + 4 = 6 + 4 = 10 \)
- \( t_5 = t_4 + 5 = 10 + 5 = 15 \)
- \( t_6 = t_5 + 6 = 15 + 6 = 21 \)
- \( t_7 = t_6 + 7 = 21 + 7 = 28 \)

3. **Generalizing the Formula**:
We can express \( t_n \) in terms of \( t_1 \) and the sum of the first \( n \) natural numbers:
\[ t_n = t_1 + 2 + 3 + \ldots + n \]
Since \( t_1 = 1 \), we can rewrite this as:
\[ t_n = 1 + (2 + 3 + \ldots + n) \]

4. **Using the Formula for the Sum of the First \( n \) Natural Numbers**:
The sum of the first \( n \) natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]
Therefore, we have:
\[ t_n = 1 + \left( \frac{n(n + 1)}{2} - 1 \right) = \frac{n(n + 1)}{2} \]

5. **Calculating \( t_{50} \)**:
Now, substituting \( n = 50 \):
\[ t_{50} = \frac{50(50 + 1)}{2} = \frac{50 \times 51}{2} \]
Simplifying this:
\[ t_{50} = \frac{2550}{2} = 1275 \]

### Final Answer:
Thus, the value of \( t_{50} \) is \( 1275 \).