The sequence `a_(1), a_(2), a_(3), ..... a_(98)` satisfies the relation `a_(n + 1) = a_(n) +1" for " n = 1, 2, 3,..... 97` and has the sum equal to 4949. Evaluate `sum_(k=1)^(49) a_(2k)`

Answer

To solve the problem, we need to analyze the sequence defined by the relation \( a_{n+1} = a_n + 1 \) for \( n = 1, 2, \ldots, 97 \) and find the sum \( \sum_{k=1}^{49} a_{2k} \).

### Step 1: Understand the Sequence
The relation \( a_{n+1} = a_n + 1 \) indicates that the sequence is an arithmetic progression (AP) where each term increases by 1 from the previous term. This means:
- \( a_2 = a_1 + 1 \)
- \( a_3 = a_1 + 2 \)
- \( a_4 = a_1 + 3 \)
- ...
- \( a_{98} = a_1 + 97 \)

### Step 2: Express the Sum of the Sequence
The sum of the first 98 terms can be expressed as:
\[ S = a_1 + (a_1 + 1) + (a_1 + 2) + \ldots + (a_1 + 97) \]
This can be simplified to:
\[ S = 98a_1 + (0 + 1 + 2 + \ldots + 97) \]
The sum of the first 97 natural numbers is given by the formula:
\[ \text{Sum} = \frac{n(n+1)}{2} = \frac{97 \times 98}{2} = 4753 \]
Thus, we have:
\[ S = 98a_1 + 4753 = 4949 \]

### Step 3: Solve for \( a_1 \)
Now we can set up the equation:
\[ 98a_1 + 4753 = 4949 \]
Subtracting 4753 from both sides gives:
\[ 98a_1 = 4949 - 4753 = 196 \]
Dividing by 98:
\[ a_1 = \frac{196}{98} = 2 \]

### Step 4: Find the Even Indexed Terms
Now we need to find \( \sum_{k=1}^{49} a_{2k} \):
- \( a_2 = a_1 + 1 = 2 + 1 = 3 \)
- \( a_4 = a_1 + 3 = 2 + 3 = 5 \)
- \( a_6 = a_1 + 5 = 2 + 5 = 7 \)
- ...
- \( a_{98} = a_1 + 97 = 2 + 97 = 99 \)

The even indexed terms form another arithmetic sequence:
\[ a_{2k} = a_1 + (2k - 1) = 2 + (2k - 1) = 2k + 1 \]
for \( k = 1, 2, \ldots, 49 \).

### Step 5: Calculate the Sum of the Even Indexed Terms
The sum can be expressed as:
\[ \sum_{k=1}^{49} a_{2k} = \sum_{k=1}^{49} (2k + 1) = \sum_{k=1}^{49} 2k + \sum_{k=1}^{49} 1 \]
Calculating each part:
1. The first sum:
\[ \sum_{k=1}^{49} 2k = 2 \sum_{k=1}^{49} k = 2 \cdot \frac{49 \cdot 50}{2} = 49 \cdot 50 = 2450 \]
2. The second sum:
\[ \sum_{k=1}^{49} 1 = 49 \]

Combining these gives:
\[ \sum_{k=1}^{49} a_{2k} = 2450 + 49 = 2499 \]

### Final Answer
Thus, the required sum \( \sum_{k=1}^{49} a_{2k} \) is:
\[ \boxed{2499} \]